:: Abstract Reduction Systems and Idea of {K}nuth {B}endix Completion
:: Algorithm
:: http://creativecommons.org/licenses/by-sa/3.0/.

environ

 vocabularies RELAT_1, XBOOLE_0, FUNCT_1, REWRITE1, TDGROUP, ABSRED_0,
      ZFMISC_1, FINSEQ_1, ARYTM_1, SUBSET_1, NUMBERS, STRUCT_0, NAT_1, ARYTM_3,
      CARD_1, XXREAL_0, ZFREFLE1, TARSKI, UNIALG_1, GROUP_1, MSUALG_6, FUNCT_2,
      INCPROJ, EQREL_1, MSUALG_1, PARTFUN1, UNIALG_2, FUNCT_4, PBOOLE, FUNCT_7,
      FINSEQ_2, FUNCOP_1, ORDINAL1, MESFUNC1;
 notations TARSKI, XBOOLE_0, ZFMISC_1, ENUMSET1, NUMBERS, XCMPLX_0, XXREAL_0,
      RELAT_1, RELSET_1, FUNCT_1, SUBSET_1, PARTFUN1, FUNCT_2, FUNCOP_1,
      EQREL_1, ORDINAL1, BINOP_1, FINSEQ_1, FINSEQ_2, NAT_1, FINSEQ_4, FUNCT_7,
      MARGREL1, STRUCT_0, PBOOLE, UNIALG_1, PUA2MSS1, REWRITE1;
 constructors RELAT_1, RELSET_1, FUNCT_2, STRUCT_0, REWRITE1, XCMPLX_0,
      XXREAL_0, NAT_1, FINSEQ_5, ENUMSET1, BINOP_1, FINSEQ_1, FINSEQ_4,
      FUNCT_7, CARD_1, XXREAL_1, UNIALG_1, PUA2MSS1, REALSET1, MARGREL1,
      EQREL_1, NUMBERS, XBOOLE_0, ZFMISC_1, SUBSET_1, FUNCT_1, PARTFUN1;
 registrations SUBSET_1, XBOOLE_0, RELSET_1, ORDINAL1, NAT_1, REWRITE1,
      XXREAL_0, XCMPLX_0, STRUCT_0, AOFA_A00, FUNCT_2, FINSEQ_1, PARTFUN1,
      FUNCOP_1, FINSEQ_2, CARD_1, MARGREL1, UNIALG_1, PUA2MSS1, RELAT_1;
 requirements BOOLE, SUBSET, NUMERALS, ARITHM, REAL;
 definitions MARGREL1, STRUCT_0, UNIALG_1, REWRITE1;
 equalities EQREL_1;
 theorems ZFMISC_1, NAT_1, FINSEQ_3, FINSEQ_5, REWRITE1, IDEA_1, XBOOLE_0,
      RELAT_1, FUNCT_1, FUNCT_2, TARSKI, SUBSET_1, ENUMSET1, SETWISEO,
      ORDINAL1, SEQ_4, MARGREL1, RELSET_1, FINSEQ_1, PARTFUN1, FINSEQ_2,
      GRFUNC_1, FUNCOP_1, FUNCT_7, PUA2MSS1, COMPUT_1;
 schemes NAT_1, RECDEF_1, RELSET_1;

begin :: Reduction and Convertibility

definition
  struct(1-sorted) ARS(#
    carrier -> set,
    reduction -> Relation of the carrier
  #);
end;

registration
  let A be non empty set, r be Relation of A;
  cluster ARS(#A, r#) -> non empty;
  coherence;
end;

registration
  cluster non empty strict for ARS;
  existence
  proof
    set A = the non empty set, r = the Relation of A;
    take X = ARS(#A,r#);
    thus X is non empty;
    thus X is strict;
  end;
end;

definition
  let X be ARS;
  let x,y be Element of X;
  pred x ==> y means [x,y] in the reduction of X;
end;

notation
  let X be ARS;
  let x,y be Element of X;
  synonym y <== x for x ==> y;
end;

definition
  let X be ARS;
  let x,y be Element of X;
  pred x =01=> y means x = y or x ==> y;
  reflexivity;
  pred x =*=> y means the reduction of X reduces x,y;
  reflexivity by REWRITE1:12;
end;

reserve X for ARS, a,b,c,u,v,w,x,y,z for Element of X;

theorem
  a ==> b implies X is non empty;

theorem Th2:
  x ==> y implies x =*=> y by REWRITE1:15;

theorem Th3:
  x =*=> y & y =*=> z implies x =*=> z by REWRITE1:16;

scheme Star{X() -> ARS, P[object]}:
  for x,y being Element of X() st x =*=> y & P[x]
  holds P[y]
provided
A1: for x,y being Element of X() st x ==> y & P[x] holds P[y]
  proof
    let x,y be Element of X();
    given p being RedSequence of the reduction of X() such that
A2: p.1 = x & p.len p = y;
    assume
A0: P[x];
    defpred Q[Nat] means $1+1 in dom p implies P[p.($1+1)];
A3: Q[0] by A0,A2;
A4: for i being Nat st Q[i] holds Q[i+1]
    proof
      let i be Nat; reconsider j = i as Element of NAT by ORDINAL1:def 12;
      assume
B1:   Q[i] & i+1+1 in dom p; then
      i+1+1 <= len p & i+1 >= 1 by NAT_1:11,FINSEQ_3:25; then
B2:   j+1 in dom p by SEQ_4:134; then
      [p.(i+1), p.(i+1+1)] in the reduction of X() by B1,REWRITE1:def 2; then
      reconsider a = p.(i+1), b = p.(i+1+1) as Element of X() by ZFMISC_1:87;
      P[a] & a ==> b by B1,B2,REWRITE1:def 2;
      hence P[p.(i+1+1)] by A1;
    end;
A5: for i being Nat holds Q[i] from NAT_1:sch 2(A3,A4);
    len p >= 0+1 by NAT_1:13; then
    (ex i being Nat st len p = 1+i) &
    len p in dom p by NAT_1:10,FINSEQ_5:6;
    hence thesis by A2,A5;
end;

scheme Star1{X() -> ARS, P[object], a, b() -> Element of X()}:
  P[b()]
provided
A1: a() =*=> b() and
A2: P[a()] and
A3: for x,y being Element of X() st x ==> y & P[x] holds P[y]
  proof
    for x,y being Element of X() st x =*=> y & P[x] holds P[y] from Star(A3);
    hence thesis by A1,A2;
  end;

scheme StarBack{X() -> ARS, P[object]}:
  for x,y being Element of X() st x =*=> y & P[y]
  holds P[x]
provided
A1: for x,y being Element of X() st x ==> y & P[y] holds P[x]
  proof
    let x,y be Element of X();
    given p being RedSequence of the reduction of X() such that
A2: p.1 = x & p.len p = y;
    assume
A0: P[y];
    defpred Q[Nat] means (len p)-$1 in dom p implies P[p.((len p)-$1)];
A3: Q[0] by A0,A2;
A4: for i being Nat st Q[i] holds Q[i+1]
    proof
      let i be Nat; assume
B1:   Q[i] & (len p)-(i+1) in dom p; then
      reconsider k = (len p)-(i+1) as Element of NAT;
B4:   k >= 0+1 by B1,FINSEQ_3:25;
      i is Element of NAT &
      k+1 = (len p)-i by ORDINAL1:def 12; then
      k+1 <= len p by IDEA_1:3; then
B2:   k in dom p & k+1 in dom p by B4,SEQ_4:134; then
      [p.k, p.(k+1)] in the reduction of X() by REWRITE1:def 2; then
      reconsider a = p.k, b = p.(k+1) as Element of X() by ZFMISC_1:87;
      P[b] & a ==> b by B1,B2,REWRITE1:def 2;
      hence thesis by A1;
    end;
A5: for i being Nat holds Q[i] from NAT_1:sch 2(A3,A4);
    len p >= 0+1 by NAT_1:13; then
    len p-1 is Nat & (len p)-((len p)-1) = 1 & 1 in dom p
    by NAT_1:21,FINSEQ_5:6;
    hence thesis by A2,A5;
end;

scheme StarBack1{X() -> ARS, P[object], a, b() -> Element of X()}:
  P[a()]
provided
A1: a() =*=> b() and
A2: P[b()] and
A3: for x,y being Element of X() st x ==> y & P[y] holds P[x]
  proof
    for x,y being Element of X() st x =*=> y & P[y] holds P[x]
    from StarBack(A3);
    hence thesis by A1,A2;
  end;

definition
  let X be ARS;
  let x,y be Element of X;
  pred x =+=> y means ex z being Element of X st x ==> z & z =*=> y;
end;

theorem Th4:
  x =+=> y iff ex z st x =*=> z & z ==> y
  proof
    thus x =+=> y implies ex z st x =*=> z & z ==> y
    proof given z such that
A1:   x ==> z & z =*=> y;
      defpred P[Element of X] means ex u st x =*=> u & u ==> $1;
A2:   for y,z st y ==> z & P[y] holds P[z]
      proof
        let y,z; assume
A3:     y ==> z;
        given u such that
A4:     x =*=> u & u ==> y;
        take y;
        u =*=> y by A4,Th2;
        hence thesis by A3,A4,Th3;
      end;
A5:   for y,z st y =*=> z & P[y] holds P[z] from Star(A2);
      thus thesis by A1,A5;
    end;
    given z such that
A6: x =*=> z & z ==> y;
    defpred P[Element of X] means ex u st $1 ==> u & u =*=> y;
A2: for y,z st y ==> z & P[z] holds P[y]
    proof
      let x,z; assume
A3:   x ==> z;
      given u such that
A4:   z ==> u & u =*=> y;
      take z;
      z =*=> u by A4,Th2;
      hence thesis by A3,A4,Th3;
    end;
A5: for y,z st y =*=> z & P[z] holds P[y] from StarBack(A2);
    thus ex z st x ==> z & z =*=> y by A5,A6;
  end;

notation
  let X,x,y;
  synonym y <=01= x for x =01=> y;
  synonym y <=*= x for x =*=> y;
  synonym y <=+= x for x =+=> y;
end;

::  x ==> y implies x =+=> y;
::  x =+=> y implies x =*=> y;
::  x =+=> y & y =*=> z implies x =+=> z;
::  x =*=> y & y =+=> z implies x =+=> z;

definition
  let X,x,y;
  pred x <==> y means x ==> y or x <== y;
  symmetry;
end;

theorem
  x <==> y iff [x,y] in (the reduction of X)\/(the reduction of X)~
  proof
A1: x ==> y iff [x,y] in the reduction of X;
A2: x <== y iff [y,x] in the reduction of X;
    [y,x] in the reduction of X iff [x,y] in (the reduction of X)~
    by RELAT_1:def 7;
    hence thesis by A1,A2,XBOOLE_0:def 3;
  end;

definition
  let X,x,y;
  pred x <=01=> y means x = y or x <==> y;
  reflexivity;
  symmetry;
  pred x <=*=> y means x,y are_convertible_wrt the reduction of X;
  reflexivity by REWRITE1:26;
  symmetry by REWRITE1:31;
end;

theorem Th6:
  x <==> y implies x <=*=> y
  proof
    assume x ==> y or x <== y;
    hence x,y are_convertible_wrt the reduction of X by REWRITE1:29,31;
  end;

theorem Th7:
  x <=*=> y & y <=*=> z implies x <=*=> z by REWRITE1:30;

scheme Star2{X() -> ARS, P[object]}:
  for x,y being Element of X() st x <=*=> y & P[x]
  holds P[y]
provided
A1:  for x,y being Element of X() st x <==> y & P[x]
  holds P[y]
proof
    let x,y be Element of X();
    set R = the reduction of X();
    assume R\/R~ reduces x,y; then :: Only 2 expansions?
::    given p being RedSequence of R\/R~ such that
    consider p being RedSequence of R\/R~ such that
A2: p.1 = x & p.len p = y by REWRITE1:def 3;
    assume
A0: P[x];
    defpred Q[Nat] means $1+1 in dom p implies P[p.($1+1)];
A3: Q[0] by A0,A2;
A4: for i being Nat st Q[i] holds Q[i+1]
    proof
      let i be Nat; reconsider j = i as Element of NAT by ORDINAL1:def 12;
      assume
B1:   Q[i] & i+1+1 in dom p; then
B4:   i+1+1 <= len p & i+1 >= 1 by NAT_1:11,FINSEQ_3:25; then
      j+1 in dom p by SEQ_4:134; then
B3:   [p.(i+1), p.(i+1+1)] in R\/R~ by B1,REWRITE1:def 2; then
      reconsider a = p.(i+1), b = p.(i+1+1) as Element of X() by ZFMISC_1:87;
      [a,b] in R or [a,b] in R~ by B3,XBOOLE_0:def 3; then
      a ==> b or b ==> a by RELAT_1:def 7; then
      P[a] & a <==> b by B1,B4,SEQ_4:134;
      hence P[p.(i+1+1)] by A1;
    end;
A5: for i being Nat holds Q[i] from NAT_1:sch 2(A3,A4);
    len p >= 0+1 by NAT_1:13; then
    (ex i being Nat st len p = 1+i) &
    len p in dom p by NAT_1:10,FINSEQ_5:6;
    hence thesis by A2,A5;
end;

scheme Star2A{X() -> ARS, P[object], a, b() -> Element of X()}:
  P[b()]
provided
A1: a() <=*=> b() and
A2: P[a()] and
A3: for x,y being Element of X() st x <==> y & P[x] holds P[y]
  proof
    for x,y being Element of X() st x <=*=> y & P[x] holds P[y] from Star2(A3);
    hence thesis by A1,A2;
  end;

definition
  let X,x,y;
  pred x <=+=> y means: Def8: ex z st x <==> z & z <=*=> y;
  symmetry
  proof let x,y;
    given z such that
A1: x <==> z & z <=*=> y;
    defpred P[Element of X] means ex u st x <=*=> u & u <==> $1;
A2: for y,z st y <==> z & P[y] holds P[z]
    proof
      let y,z; assume
A3:   y <==> z;
      given u such that
A4:   x <=*=> u & u <==> y;
      take y;
      u <=*=> y by A4,Th6;
      hence thesis by A3,A4,Th7;
    end;
A5: for y,z st y <=*=> z & P[y] holds P[z] from Star2(A2);
    ex u st x <=*=> u & u <==> y by A1,A5;
    hence thesis;
  end;
end;

theorem Th8:
  x <=+=> y iff ex z st x <=*=> z & z <==> y
  proof
    x <=+=> y iff ex z st y <==> z & z <=*=> x by Def8;
    hence thesis;
  end;

theorem Lem1:
  x =01=> y implies x =*=> y by Th2;

theorem Lem2:
  x =+=> y implies x =*=> y
  proof
    assume
A1: x =+=> y;
    consider z such that
A2: x ==> z & z =*=> y by A1;
A3: x =*=> z by A2,Th2;
    thus x =*=> y by A2,A3,Th3;
  end;

theorem
  x ==> y implies x =+=> y;

theorem Lem3:
  x ==> y & y ==> z implies x =*=> z
  proof
    assume
A1: x ==> y;
    assume
A2: y ==> z;
A3: x =*=> y by A1,Th2;
A4: y =*=> z by A2,Th2;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem Lem4:
  x ==> y & y =01=> z implies x =*=> z
  proof
    assume
A1: x ==> y;
    assume
A2: y =01=> z;
A3: x =*=> y by A1,Th2;
A4: y =*=> z by A2,Lem1;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem Lem5:
  x ==> y & y =*=> z implies x =*=> z
  proof
    assume
A1: x ==> y;
    assume
A2: y =*=> z;
A3: x =*=> y by A1,Th2;
    thus x =*=> z by A3,A2,Th3;
  end;

theorem Lem5A:
  x ==> y & y =+=> z implies x =*=> z
  proof
    assume
A1: x ==> y;
    assume
A2: y =+=> z;
A3: x =*=> y by A1,Th2;
A4: y =*=> z by A2,Lem2;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem
  x =01=> y & y ==> z implies x =*=> z
  proof
    assume
A1: x =01=> y;
    assume
A2: y ==> z;
A3: x =*=> y by A1,Lem1;
A4: y =*=> z by A2,Th2;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem
  x =01=> y & y =01=> z implies x =*=> z
  proof
    assume
A1: x =01=> y;
    assume
A2: y =01=> z;
A3: x =*=> y by A1,Lem1;
A4: y =*=> z by A2,Lem1;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem Lem8:
  x =01=> y & y =*=> z implies x =*=> z
  proof
    assume
A1: x =01=> y;
    assume
A2: y =*=> z;
A3: x =*=> y by A1,Lem1;
    thus x =*=> z by A3,A2,Th3;
  end;

theorem
  x =01=> y & y =+=> z implies x =*=> z
  proof
    assume
A1: x =01=> y;
    assume
A2: y =+=> z;
A3: x =*=> y by A1,Lem1;
A4: y =*=> z by A2,Lem2;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem Lem10:
  x =*=> y & y ==> z implies x =*=> z
  proof
    assume
A1: x =*=> y;
    assume
A2: y ==> z;
A4: y =*=> z by A2,Th2;
    thus x =*=> z by A1,A4,Th3;
  end;

theorem Lem11:
  x =*=> y & y =01=> z implies x =*=> z
  proof
    assume
A1: x =*=> y;
    assume
A2: y =01=> z;
A4: y =*=> z by A2,Lem1;
    thus x =*=> z by A1,A4,Th3;
  end;

theorem Lem11A:
  x =*=> y & y =+=> z implies x =*=> z
  proof
    assume
A1: x =*=> y;
    assume
A2: y =+=> z;
A4: y =*=> z by A2,Lem2;
    thus x =*=> z by A1,A4,Th3;
  end;

theorem
  x =+=> y & y ==> z implies x =*=> z
  proof
    assume
A1: x =+=> y;
    assume
A2: y ==> z;
A3: x =*=> y by A1,Lem2;
A4: y =*=> z by A2,Th2;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem
  x =+=> y & y =01=> z implies x =*=> z
  proof
    assume
A1: x =+=> y;
    assume
A2: y =01=> z;
A3: x =*=> y by A1,Lem2;
A4: y =*=> z by A2,Lem1;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem
  x =+=> y & y =+=> z implies x =*=> z
  proof
    assume
A1: x =+=> y;
    assume
A2: y =+=> z;
A3: x =*=> y by A1,Lem2;
A4: y =*=> z by A2,Lem2;
    thus x =*=> z by A3,A4,Th3;
  end;

theorem
  x ==> y & y ==> z implies x =+=> z by Th2;


theorem
  x ==> y & y =01=> z implies x =+=> z by Lem1;

theorem
  x ==> y & y =+=> z implies x =+=> z by Lem2;

theorem
  x =01=> y & y ==> z implies x =+=> z by Lem1,Th4;

theorem
  x =01=> y & y =+=> z implies x =+=> z
  proof
    assume
A1: x =01=> y;
    assume
A2: y =+=> z;
    consider u such that
A3: y =*=> u & u ==> z by A2,Th4;
    thus x =+=> z by A3,A1,Lem8,Th4;
  end;

theorem
  x =*=> y & y =+=> z implies x =+=> z
  proof
    assume
A1: x =*=> y;
    assume
A2: y =+=> z;
    consider u such that
A3: y =*=> u & u ==> z by A2,Th4;
    thus x =+=> z by A3,A1,Th3,Th4;
  end;

theorem
  x =+=> y & y ==> z implies x =+=> z by Lem10;

theorem
  x =+=> y & y =01=> z implies x =+=> z by Lem11;

theorem
  x =+=> y & y =*=> z implies x =+=> z by Th3;

theorem
  x =+=> y & y =+=> z implies x =+=> z by Lem11A;

theorem Lem1A:
  x <=01=> y implies x <=*=> y by Th6;

theorem Lem2A:
  x <=+=> y implies x <=*=> y
  proof
    assume
A1: x <=+=> y;
    consider z such that
A2: x <==> z & z <=*=> y by A1;
A3: x <=*=> z by A2,Th6;
    thus x <=*=> y by A2,A3,Th7;
  end;

theorem LemB:
  x <==> y implies x <=+=> y;

theorem
  x <==> y & y <==> z implies x <=*=> z
  proof
    assume
A1: x <==> y;
    assume
A2: y <==> z;
A3: x <=*=> y by A1,Th6;
A4: y <=*=> z by A2,Th6;
    thus x <=*=> z by A3,A4,Th7;
  end;

theorem Lem4A:
  x <==> y & y <=01=> z implies x <=*=> z
  proof
    assume
A1: x <==> y;
    assume
A2: y <=01=> z;
A3: x <=*=> y by A1,Th6;
A4: y <=*=> z by A2,Lem1A;
    thus x <=*=> z by A3,A4,Th7;
  end;

theorem
  x <=01=> y & y <==> z implies x <=*=> z by Lem4A;

theorem Lem5a:
  x <==> y & y <=*=> z implies x <=*=> z
  proof
    assume
A1: x <==> y;
    assume
A2: y <=*=> z;
A3: x <=*=> y by A1,Th6;
    thus x <=*=> z by A3,A2,Th7;
  end;

theorem
  x <=*=> y & y <==> z implies x <=*=> z by Lem5a;

theorem Lem5B:
  x <==> y & y <=+=> z implies x <=*=> z
  proof
    assume
A1: x <==> y;
    assume
A2: y <=+=> z;
A3: x <=*=> y by A1,Th6;
A4: y <=*=> z by A2,Lem2A;
    thus x <=*=> z by A3,A4,Th7;
  end;

theorem
  x <=+=> y & y <==> z implies x <=*=> z by Lem5B;

theorem
  x <=01=> y & y <=01=> z implies x <=*=> z
  proof
    assume
A1: x <=01=> y;
    assume
A2: y <=01=> z;
A3: x <=*=> y by A1,Lem1A;
A4: y <=*=> z by A2,Lem1A;
    thus x <=*=> z by A3,A4,Th7;
  end;

theorem Lm8:
  x <=01=> y & y <=*=> z implies x <=*=> z
  proof
    assume
A1: x <=01=> y;
    assume
A2: y <=*=> z;
A3: x <=*=> y by A1,Lem1A;
    thus x <=*=> z by A3,A2,Th7;
  end;

theorem
  x <=*=> y & y <=01=> z implies x <=*=> z by Lm8;

theorem Lem9:
  x <=01=> y & y <=+=> z implies x <=*=> z
  proof
    assume
A1: x <=01=> y;
    assume
A2: y <=+=> z;
A3: x <=*=> y by A1,Lem1A;
A4: y <=*=> z by A2,Lem2A;
    thus x <=*=> z by A3,A4,Th7;
  end;

theorem
  x <=+=> y & y <=01=> z implies x <=*=> z by Lem9;

theorem Lem11A:
  x <=*=> y & y <=+=> z implies x <=*=> z
  proof
    assume
A1: x <=*=> y;
    assume
A2: y <=+=> z;
A4: y <=*=> z by A2,Lem2A;
    thus x <=*=> z by A1,A4,Th7;
  end;

theorem
  x <=+=> y & y <=+=> z implies x <=*=> z
  proof
    assume
A1: x <=+=> y;
    assume
A2: y <=+=> z;
A3: x <=*=> y by A1,Lem2A;
A4: y <=*=> z by A2,Lem2A;
    thus x <=*=> z by A3,A4,Th7;
  end;

theorem
  x <==> y & y <==> z implies x <=+=> z by Th6;

theorem
  x <==> y & y <=01=> z implies x <=+=> z by Lem1A;

theorem
  x <==> y & y <=+=> z implies x <=+=> z by Lem2A;

theorem Lem18:
  x <=01=> y & y <=+=> z implies x <=+=> z
  proof
    assume
A1: x <=01=> y;
    assume
A2: y <=+=> z;
    consider u such that
A3: y <=*=> u & u <==> z by A2,Th8;
    thus x <=+=> z by A3,A1,Lm8,Th8;
  end;

theorem
  x <=*=> y & y <=+=> z implies x <=+=> z
  proof
    assume
A1: x <=*=> y;
    assume
A2: y <=+=> z;
    consider u such that
A3: y <=*=> u & u <==> z by A2,Th8;
    thus x <=+=> z by A3,A1,Th7,Th8;
  end;

theorem
  x <=+=> y & y <=+=> z implies x <=+=> z by Lem11A;

theorem Lem31:
  x <=01=> y implies x <== y or x = y or x ==> y
  proof
    assume
A1: x <=01=> y;
A2: x <==> y or x = y by A1;
    thus x <== y or x = y or x ==> y by A2;
  end;

theorem
  x <== y or x = y or x ==> y implies x <=01=> y
  proof
    assume
A1: x <== y or x = y or x ==> y;
A2: x <==> y or x = y by A1;
    thus x <=01=> y by A2;
  end;

theorem
  x <=01=> y implies x <=01= y or x ==> y
  proof
    assume
A1: x <=01=> y;
A2: x <==> y or x = y by A1;
    thus x <=01= y or x ==> y by A2;
  end;

theorem
  x <=01= y or x ==> y implies x <=01=> y
  proof
    assume
A1: x <=01= y or x ==> y;
A3: x <==> y or x = y by A1;
    thus x <=01=> y by A3;
  end;

theorem
  x <=01=> y implies x <=01= y or x =+=> y
  proof
    assume
A1: x <=01=> y;
A2: x <==> y or x = y by A1;
    thus x <=01= y or x =+=> y by A2;
  end;

theorem
  x <=01=> y implies x <=01= y or x <==> y;

theorem
  x <=01= y or x <==> y implies x <=01=> y
  proof
    assume
A1: x <=01= y or x <==> y;
A3: x = y or x <==> y by A1;
    thus x <=01=> y by A3;
  end;

theorem
  x <=*=> y & y ==> z implies x <=+=> z
  proof
    assume
A1: x <=*=> y;
    assume
A2: y ==> z;
A4: y <==> z by A2;
    thus x <=+=> z by A1,A4,Def8;
  end;

theorem
  x <=+=> y & y ==> z implies x <=+=> z
  proof
    assume
A1: x <=+=> y;
    assume
A2: y ==> z;
A3: x <=*=> y by A1,Lem2A;
A4: y <==> z by A2;
    thus x <=+=> z by A3,A4,Def8;
  end;

theorem
  x <=01=> y implies x <=01= y or x ==> y
  proof
    assume
A1: x <=01=> y;
A2: x = y or x <==> y by A1;
    thus x <=01= y or x ==> y by A2;
  end;

theorem
x <=01=> y implies x <=01= y or x =+=> y
  proof
    assume
A1: x <=01=> y;
A2: x = y or x <==> y by A1;
    thus x <=01= y or x =+=> y by A2;
  end;

theorem Lem43:
  x <=01= y or x ==> y implies x <=01=> y
  proof
    assume
A1: x <=01= y or x ==> y;
A3: x <==> y or x = y by A1;
    thus x <=01=> y by A3;
  end;

theorem
  x <=01= y or x <==> y implies x <=01=> y
  proof
    assume
A1: x <=01= y or x <==> y;
A3: x <==> y or x = y by A1;
    thus x <=01=> y by A3;
  end;

theorem
  x <=01=> y implies x <=01= y or x <==> y;

theorem
  x <=+=> y & y ==> z implies x <=+=> z
  proof
    assume
A1: x <=+=> y;
    assume
A2: y ==> z;
A3: x <=*=> y by A1,Lem2A;
A4: y <==> z by A2;
    thus x <=+=> z by A3,A4,Def8;
  end;

theorem
  x <=*=> y & y ==> z implies x <=+=> z
  proof
    assume
A1: x <=*=> y;
    assume
A2: y ==> z;
A4: y <==> z by A2;
    thus x <=+=> z by A1,A4,Def8;
  end;

theorem
  x <=01=> y & y ==> z implies x <=+=> z
  proof
    assume
A1: x <=01=> y;
    assume
A2: y ==> z;
A4: y <==> z by A2;
    thus x <=+=> z by A1,A4,Lem1A,Def8;
  end;

theorem
  x <=+=> y & y =01=> z implies x <=+=> z
  proof
    assume
A1: x <=+=> y;
    assume
A2: y =01=> z;
A3: y <=01=> z by A2,Lem43;
    thus x <=+=> z by A1,A3,Lem18;
  end;

theorem
  x <==> y & y =01=> z implies x <=+=> z
  proof
    assume
A1: x <==> y;
    assume
A2: y =01=> z;
A3: y <=01=> z by A2,Lem43;
    thus x <=+=> z by A3,A1,LemB,Lem18;
  end;

theorem
  x ==> y & y ==> z & z ==> u implies x =+=> u by Lem3;

theorem
  x ==> y & y =01=> z & z ==> u implies x =+=> u by Lem4,Th4;

theorem
  x ==> y & y =*=> z & z ==> u implies x =+=> u by Lem5,Th4;

theorem
  x ==> y & y =+=> z & z ==> u implies x =+=> u
  proof
    assume
A1: x ==> y;
    assume
A2: y =+=> z;
    assume
A3: z ==> u;
A4: x =*=> z by A1,A2,Lem5A;
    thus x =+=> u by A3,A4,Th4;
  end;

theorem LemZ:
  x =*=> y implies x <=*=> y
  proof
    assume
A1: x =*=> y;
    defpred P[Element of X] means x <=*=> $1;
A2: P[x];
A3: for y,z st y ==> z & P[y] holds P[z]
    proof
      let y,z;
      assume
A4:   y ==> z;
      assume
A5:   P[y];
A6:   y <==> z by A4;
A7:   y <=*=> z by A6,Th6;
      thus P[z] by A5,A7,Th7;
    end;
    thus P[y] from Star1(A1,A2,A3);
  end;

theorem
  for z st
    for x,y st x ==> z & x ==> y holds y ==> z
  for x,y st x ==> z & x =*=> y
  holds y ==> z
  proof
    let z;
    assume
A:  for x,y st x ==> z & x ==> y holds y ==> z;
    let x,y;
    assume
B:  x ==> z & x =*=> y;
    defpred P[Element of X] means $1 ==> z;
C:  for u,v st u ==> v & P[u] holds P[v] by A;
D:  for u,v st u =*=> v & P[u] holds P[v] from Star(C);
    thus y ==> z by B,D;
end;

theorem
  (for x,y st x ==> y holds y ==> x)
  implies
  for x,y st x <=*=> y holds x =*=> y
  proof
    assume
A:  for x,y st x ==> y holds y ==> x;
    let x,y;
    assume
B:  x <=*=> y;
    defpred P[Element of X] means x =*=> $1;
C:  for u,v st u <==> v & P[u] holds P[v] by A,Lem10;
D:  for u,v st u <=*=> v & P[u] holds P[v] from Star2(C);
    thus x =*=> y by B,D;
end;

theorem LemN:
  x =*=> y implies x = y or x =+=> y
  proof
    assume
A1: x =*=> y;
    defpred P[Element of X] means x = $1 or x =+=> $1;
A2: P[x];
A3: for y,z st y ==> z & P[y] holds P[z]
    proof
      let y,z;
      assume
A4:   y ==> z;
      assume
A5:   P[y];
A6:   x =*=> y by A5,Lem2;
      thus P[z] by A6,A4,Th4;
    end;
    thus P[y] from Star1(A1,A2,A3);
  end;

theorem
  (for x,y,z st x ==> y & y ==> z holds x ==> z)
  implies
  for x,y st x =+=> y holds x ==> y
  proof
    assume
A1: for x,y,z st x ==> y & y ==> z holds x ==> z;
    let x,y;
    assume
A2: x =+=> y;
    consider z such that
A3: x ==> z and
A4: z =*=> y by A2;
    defpred P[Element of X] means x ==> $1;
A5: P[z] by A3;
A6: for u,v st u ==> v & P[u] holds P[v] by A1;
    thus P[y] from Star1(A4,A5,A6);
  end;

begin :: Examples of ARS

scheme ARSex{A() -> non empty set, R[object,object]}:
  ex X being strict non empty ARS st the carrier of X = A() &
  for x,y being Element of X holds x ==> y iff R[x,y]
  proof
    consider r being Relation of A() such that
A1: for x,y being Element of A() holds [x,y] in r iff R[x,y]
    from RELSET_1:sch 2;
    take X = ARS(#A(), r#);
    thus the carrier of X = A();
    thus thesis by A1;
  end;

definition
  func ARS_01 -> strict ARS means:
Def18:
  the carrier of it = {0,1} &
  the reduction of it = [:{0},{0,1}:];
  existence
  proof
    {0} c= {0,1} by ZFMISC_1:7; then
    reconsider r = [:{0},{0,1}:] as Relation of {0,1} by ZFMISC_1:96;
    take X = ARS(#{0,1}, r#);
    thus thesis;
  end;
  uniqueness;
  func ARS_02 -> strict ARS means:
Def19:
  the carrier of it = {0,1,2} &
  the reduction of it = [:{0},{0,1,2}:];
  existence
  proof
    {0} c= {0,1,2} by SETWISEO:1; then
    reconsider r = [:{0},{0,1,2}:] as Relation of {0,1,2} by ZFMISC_1:96;
    take X = ARS(#{0,1,2}, r#);
    thus thesis;
  end;
  uniqueness;
end;

registration
  cluster ARS_01 -> non empty;
  coherence by Def18;
  cluster ARS_02 -> non empty;
  coherence by Def19;
end;

reserve i,j,k for Element of ARS_01;

theorem ThA1:
  for s being set holds s is Element of ARS_01 iff s = 0 or s = 1
  proof
    let s be set;
    the carrier of ARS_01 = {0,1} by Def18;
    hence thesis by TARSKI:def 2;
  end;

theorem
  i ==> j iff i = 0
  proof
    the reduction of ARS_01 = [:{0},{0,1}:] by Def18; then
    i ==> j iff i in {0} & j in {0,1} by ZFMISC_1:87; then
    i ==> j iff i = 0 & (j = 0 or j = 1) by TARSKI:def 1,def 2;
    hence thesis by ThA1;
  end;

reserve l,m,n for Element of ARS_02;

theorem ThB1:
  for s being set holds s is Element of ARS_02 iff s = 0 or s = 1 or s = 2
  proof
    let s be set;
    the carrier of ARS_02 = {0,1,2} by Def19;
    hence thesis by ENUMSET1:def 1;
  end;

theorem
  m ==> n iff m = 0
  proof
    the reduction of ARS_02 = [:{0},{0,1,2}:] by Def19; then
    m ==> n iff m in {0} & n in {0,1,2} by ZFMISC_1:87; then
    m ==> n iff m = 0 & (n = 0 or n = 1 or n = 2)
    by TARSKI:def 1,ENUMSET1:def 1;
    hence thesis by ThB1;
  end;

begin :: Normal Forms

definition
  let X,x;
  attr x is normform means not ex y st x ==> y;
end;

theorem Ch1:
  x is normform iff x is_a_normal_form_wrt the reduction of X
  proof set R = the reduction of X;
    thus x is normform implies x is_a_normal_form_wrt the reduction of X
    proof assume
Z0:   not ex y st x ==> y;
      let a be object;
      assume
Z1:   [x,a] in the reduction of X; then
      reconsider y = a as Element of X by ZFMISC_1:87;
      x ==> y by Z1;
      hence thesis by Z0;
    end;
    assume
Z1: not ex b being object st [x,b] in R;
    let y;
    assume [x,y] in the reduction of X;
    hence thesis by Z1;
  end;

definition
  let X,x,y;
  pred x is_normform_of y means x is normform & y =*=> x;
end;

theorem Ch2:
  x is_normform_of y iff x is_a_normal_form_of y, the reduction of X
  proof set R = the reduction of X;
    thus x is_normform_of y implies x is_a_normal_form_of y, R
    proof assume
      x is normform & R reduces y,x;
      hence x is_a_normal_form_wrt R & R reduces y,x by Ch1;
    end;
    assume x is_a_normal_form_wrt R & R reduces y,x;
    hence x is normform & R reduces y,x by Ch1;
  end;

definition
  let X,x;
  attr x is normalizable means ex y st y is_normform_of x;
end;

theorem Ch3:
  x is normalizable iff x has_a_normal_form_wrt the reduction of X
  proof
    set R = the reduction of X;
A0: field R c= (the carrier of X)\/the carrier of X by RELSET_1:8;
    thus x is normalizable implies x has_a_normal_form_wrt R
    proof
      given y such that
A1:   y is_normform_of x;
      take y; thus thesis by A1,Ch2;
    end;
    given a being object such that
A2: a is_a_normal_form_of x, R;
    R reduces x,a by A2,REWRITE1:def 6; then
    x = a or a in field R by REWRITE1:18; then
    reconsider a as Element of X by A0;
    take a; thus thesis by A2,Ch2;
  end;

definition
  let X;
  attr X is WN means for x holds x is normalizable;
  attr X is SN means
  for f being Function of NAT, the carrier of X
  ex i being Nat st not f.i ==> f.(i+1);
  attr X is UN* means
  for x,y,z st y is_normform_of x & z is_normform_of x holds y = z;
  attr X is UN means
  for x,y st x is normform & y is normform & x <=*=> y holds x = y;
  attr X is N.F. means
  for x,y st x is normform & x <=*=> y holds y =*=> x;
end;

theorem
  X is WN iff the reduction of X is weakly-normalizing
  proof set R = the reduction of X;
A0: field R c= (the carrier of X)\/the carrier of X by RELSET_1:8;
    thus X is WN implies R is weakly-normalizing
    proof assume
A1:   for x holds x is normalizable;
      let a be object; assume a in field R; then
      reconsider a as Element of X by A0;
      a is normalizable by A1;
      hence thesis by Ch3;
    end;
    assume
A2: for a being object st a in field R
    holds a has_a_normal_form_wrt R;
    let x;
    per cases;
    suppose
      x in field R;
      hence thesis by A2,Ch3;
    end;
    suppose
A3:   not x in field R;
      take x;
      thus x is normform
      proof
        let y;
        thus not [x,y] in R by A3,RELAT_1:15;
      end;
      thus thesis;
    end;
  end;

theorem Ch7:
  X is SN implies the reduction of X is strongly-normalizing
  proof set R = the reduction of X;
    set A = the carrier of X;
A0: field R c= A \/ A by RELSET_1:8;
    assume
A1: for f being Function of NAT, A
    ex i being Nat st not f.i ==> f.(i+1);
    let f be ManySortedSet of NAT;
    per cases;
    suppose f is A-valued; then
      rng f c= A & dom f = NAT by RELAT_1:def 19,PARTFUN1:def 2; then
      reconsider g = f as Function of NAT, A by FUNCT_2:2;
      consider i being Nat such that
A2:   not g.i ==> g.(i+1) by A1;
      take i;
      thus not [f.i,f.(i+1)] in R by A2;
    end;
    suppose
      f is not A-valued; then
      consider a being object such that
A3:   a in rng f & not a in A by TARSKI:def 3,RELAT_1:def 19;
      consider i being object such that
A4:   i in dom f & a = f.i by A3,FUNCT_1:def 3;
      reconsider i as Element of NAT by A4;
      take i;
      assume [f.i,f.(i+1)] in R; then
      a in field R by A4,RELAT_1:15;
      hence thesis by A0,A3;
    end;
  end;

theorem Ch8:
  X is non empty & the reduction of X is strongly-normalizing implies X is SN
  proof set R = the reduction of X;
    set A = the carrier of X;
    assume
A1: X is non empty;
    assume
A5: for f being ManySortedSet of NAT
    ex i being Nat st not [f.i,f.(i+1)] in R;
    let f be Function of NAT, A;
    consider i being Nat such that
A6: not [f.i,f.(i+1)] in R by A1,A5;
    take i;
    thus not [f.i,f.(i+1)] in R by A6;
  end;

reserve A for set;

theorem ThSN:
  for X holds X is SN iff
  not ex A,z st z in A & for x st x in A ex y st y in A & x ==> y
  proof
    let X;
    thus X is SN implies not ex A,z st z in A &
    for x st x in A ex y st y in A & x ==> y
    proof assume
00:   for f being Function of NAT, the carrier of X
      ex i being Nat st not f.i ==> f.(i+1);
      given A,z such that
01:   z in A & for x st x in A ex y st y in A & x ==> y;
      ex y st y in A & z ==> y by 01; then
      reconsider X0 = X as non empty ARS;
      reconsider z0 = z as Element of X0;
      defpred P[Nat,Element of X0,Element of X0] means
      $2 in A implies $3 in A & $2 ==> $3;
02:   for i being Nat, x being Element of X0
      ex y being Element of X0 st P[i,x,y] by 01;
      consider f being Function of NAT, the carrier of X0 such that
03:   f.0 = z0 and
04:   for i being Nat holds P[i,f.i,f.(i+1)]
      from RECDEF_1:sch 2(02);
      defpred Q[Nat] means f.$1 ==> f.($1+1) & f.$1 in A;
05:   Q[0] by 01,03,04;
06:   now let i be Nat; assume Q[i]; then
        f.(i+1) in A by 04;
        hence Q[i+1] by 04;
      end;
      for i being Nat holds Q[i] from NAT_1:sch 2(05,06);
      hence contradiction by 00;
    end;
    assume
00: not ex A,z st z in A &
    for x st x in A ex y st y in A & x ==> y;
    given f being Function of NAT, the carrier of X such that
01: for i being Nat holds f.i ==> f.(i+1);
    f.0 ==> f.(0+1) by 01; then
04: X is non empty & 0 in NAT by ORDINAL1:def 12; then
02: f.0 in rng f by FUNCT_2:4;
    now let x; assume x in rng f; then
      consider i being object such that
03:   i in dom f & x = f.i by FUNCT_1:def 3;
      reconsider i as Element of NAT by 03;
      take y = f.(i+1);
      thus y in rng f by 04,FUNCT_2:4;
      thus x ==> y by 01,03;
    end;
    hence contradiction by 00,02;
  end;

scheme notSN{X() -> ARS, P[object]}:
  X() is not SN
provided
A1: ex x being Element of X() st P[x]
 and
A2: for x being Element of X() st P[x]
    ex y being Element of X() st P[y] & x ==> y
  proof
    set A = {x where x is Element of X(): P[x]};
    consider z being Element of X() such that
A3: P[z] by A1;
A4: z in A by A3;
    now let x be Element of X(); assume x in A; then
      ex a being Element of X() st x = a & P[a]; then
      consider y being Element of X() such that
A6:   P[y] & x ==> y by A2;
      take y; thus y in A by A6;
      thus x ==> y by A6;
    end;
    hence thesis by A4,ThSN;
  end;

theorem
  X is UN iff the reduction of X is with_UN_property
  proof
    set R = the reduction of X;
    set A = the carrier of X;
A0: field R c= A \/ A by RELSET_1:8;
    thus X is UN implies R is with_UN_property
    proof
      assume
A1:   for x,y st x is normform & y is normform & x <=*=> y holds x = y;
      let a,b be object;
      assume
A2:   a is_a_normal_form_wrt R & b is_a_normal_form_wrt R &
      a,b are_convertible_wrt R;
      per cases;
      suppose a in A & b in A; then
        reconsider x = a, y = b as Element of X;
        x is normform & y is normform & x <=*=> y by A2,Ch1;
        hence a = b by A1;
      end;
      suppose not a in A or not b in A; then
        not a in field R or not b in field R by A0;
        hence a = b by A2,REWRITE1:28,31;
      end;
    end;
    assume
A4: for a,b being object
    st a is_a_normal_form_wrt R & b is_a_normal_form_wrt R &
    a,b are_convertible_wrt R holds a = b;
    let x,y; assume
    x is normform & y is normform & x <=*=> y; then
    x is_a_normal_form_wrt R & y is_a_normal_form_wrt R &
    x,y are_convertible_wrt R by Ch1;
    hence x = y by A4;
  end;

theorem
  X is N.F. iff the reduction of X is with_NF_property
  proof
    set R = the reduction of X;
    set A = the carrier of X;
A0: field R c= A \/ A by RELSET_1:8;
    thus X is N.F. implies R is with_NF_property
    proof
      assume
A1:   for x,y st x is normform & x <=*=> y holds y =*=> x;
      let a,b be object; assume
A2:   a is_a_normal_form_wrt R & a,b are_convertible_wrt R;
      per cases;
      suppose a in A & b in A; then
        reconsider x = a, y = b as Element of X;
        x is normform & x <=*=> y by A2,Ch1; then
        y =*=> x by A1;
        hence R reduces b,a;
      end;
      suppose not a in A or not b in A; then
        not a in field R or not b in field R by A0; then
        a = b by A2,REWRITE1:28,31;
        hence R reduces b,a by REWRITE1:12;
      end;
    end;
    assume
B1: for a,b being object st
      a is_a_normal_form_wrt R & a,b are_convertible_wrt R
    holds R reduces b,a;
    let x,y; assume
    x is normform & x <=*=> y;
    hence R reduces y,x by B1,Ch1;
  end;

definition
  let X;
  let x such that
A: ex y st y is_normform_of x and
B: for y,z st y is_normform_of x & z is_normform_of x holds y = z;
  func nf x -> Element of X means:
Def17:
  it is_normform_of x;
  existence by A;
  uniqueness by B;
end;

theorem
  (ex y st y is_normform_of x) &
  (for y,z st y is_normform_of x & z is_normform_of x holds y = z)
  implies nf x = nf(x, the reduction of X)
  proof set R = the reduction of X;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    given y such that
A0: y is_normform_of x;
B0: x has_a_normal_form_wrt R by A0,Ch2,REWRITE1:def 11;
    assume
A1: for y,z st y is_normform_of x & z is_normform_of x holds y = z; then
    nf x is_normform_of x by A0,Def17; then
A2: nf x is_a_normal_form_of x,R by Ch2;
    now
      let b,c be object; assume
A3:   b is_a_normal_form_of x,R & c is_a_normal_form_of x,R; then
A4:   R reduces x,b & R reduces x,c by REWRITE1:def 6;
      per cases;
      suppose x in field R; then
        b in field R & c in field R by A4,REWRITE1:19; then
        reconsider y = b, z = c as Element of X by F0;
        y is_normform_of x & z is_normform_of x by A3,Ch2;
        hence b = c by A1;
      end;
      suppose not x in field R; then
        x = b & x = c by A4,REWRITE1:18;
        hence b = c;
      end;
    end;
    hence nf x = nf(x, the reduction of X) by B0,A2,REWRITE1:def 12;
 end;

theorem LemN1:
  x is normform & x =*=> y implies x = y
  proof
    assume
A1: x is normform;
    assume
A2: x =*=> y;
A4: not x =+=> y by A1;
    thus x = y by A2,A4,LemN;
  end;

theorem LemN2:
  x is normform implies x is_normform_of x;

theorem
  x is normform & y ==> x implies x is_normform_of y by Th2;

theorem
  x is normform & y =01=> x implies x is_normform_of y by Lem1;

theorem
  x is normform & y =+=> x implies x is_normform_of y by Lem2;

theorem
  x is_normform_of y & y is_normform_of x implies x = y by LemN1;

theorem LemN6:
  x is_normform_of y & z ==> y implies x is_normform_of z by Lem5;

theorem LemN7:
  x is_normform_of y & z =*=> y implies x is_normform_of z by Th3;

theorem
  x is_normform_of y & z =*=> x implies x is_normform_of z;

registration
  let X;
  cluster normform -> normalizable for Element of X;
  coherence
  proof let x;
    assume
A1: x is normform;
    take x;
    thus x is_normform_of x by A1;
  end;
end;

theorem LemN5:
  x is normalizable & y ==> x implies y is normalizable by LemN6;

theorem ThWN1:
  X is WN iff for x ex y st y is_normform_of x
  proof
    thus X is WN implies for x ex y st y is_normform_of x
    proof
      assume
A1:   for x holds x is normalizable;
      let x;
A2:   x is normalizable by A1;
      thus ex y st y is_normform_of x by A2;
    end;
    assume
A3: for x ex y st y is_normform_of x;
    let x; thus ex y st y is_normform_of x by A3;
  end;

theorem
  (for x holds x is normform) implies X is WN
  proof
    assume
A1: for x holds x is normform;
    let x;
A2: x is normform by A1;
    thus ex y st y is_normform_of x by A2,LemN2;
  end;

registration
  cluster SN -> WN for ARS;
  coherence
  proof let X;
    assume
A1: X is SN;
    assume
A2: X is not WN;
    consider z such that
A3: z is not normalizable by A2;
    set A = {x: x is not normalizable};
A4: z in A by A3;
A5: for x st x in A ex y st y in A & x ==> y
    proof
      let x;
      assume x in A; then
A6:   ex y st x = y & y is not normalizable; then
      x is not normform; then
      consider y such that
A7:   x ==> y;
      take y;
      y is not normalizable by A6,A7,LemN5;
      hence y in A;
      thus x ==> y by A7;
    end;
    thus contradiction by A1,A4,A5,ThSN;
  end;
end;

theorem LmA:
  x <> y & (for a,b holds a ==> b iff a = x)
  implies y is normform & x is normalizable
  proof
    assume
Z0: x <> y;
    assume
Z2: for a,b holds a ==> b iff a = x;
    thus y is normform by Z0,Z2;
    take y; thus y is normform by Z0,Z2;
    thus thesis by Z2,Th2;
  end;

theorem
  ex X st X is WN & X is not SN
  proof
    defpred R[set,set] means $1 = 0;
    consider X being strict non empty ARS such that
A1: the carrier of X = {0,1} and
A2: for x,y being Element of X holds x ==> y iff R[x,y] from ARSex;
    reconsider z = 0, o = 1 as Element of X by A1,TARSKI:def 2;
A3: z <> o;
    take X;
    thus X is WN
    proof
      let x be Element of X;
      x = 0 or x = 1 by A1,TARSKI:def 2; then
      x is normform or x is normalizable by A2,A3,LmA;
      hence thesis;
    end;
    set A = {z};
A4: z in A by TARSKI:def 1;
    now
      let x be Element of X;
      assume x in A; then
A5:   x = z by TARSKI:def 1;
      take y = z;
      thus y in A & x ==> y by A2,A5,TARSKI:def 1;
    end;
    hence X is not SN by A4,ThSN;
  end;

registration
  cluster N.F. -> UN* for ARS;
  coherence
  proof let X;
    assume
A1: for x,y st x is normform & x <=*=> y holds y =*=> x;
    let x,y,z;
    assume
A2: y is normform & x =*=> y;
    assume
A3: z is normform & x =*=> z;
A4: x <=*=> y & x <=*=> z by A2,A3,LemZ;
A5: y <=*=> z by A4,Th7;
    thus y = z by A2,A1,A3,A5,LemN1;
  end;

  cluster N.F. -> UN for ARS;
  coherence by LemN1;

  cluster UN -> UN* for ARS;
  coherence
  proof let X;
    assume
A1: for x,y st x is normform & y is normform & x <=*=> y holds x = y;
    let x,y,z;
    assume
A2: y is normform & x =*=> y;
    assume
A3: z is normform & x =*=> z;
A4: x <=*=> y & x <=*=> z by A2,A3,LemZ;
    thus y = z by A1,A2,A3,A4,Th7;
  end;
end;

theorem LemN12:
  X is WN UN* & x is normform & x <=*=> y implies y =*=> x
  proof
    assume
A1: X is WN UN*;
    assume
A2: x is normform;
    assume
A3: x <=*=> y;
    defpred P[Element of X] means $1 =*=> x;
A4: for y,z st y <==> z & P[y] holds P[z]
    proof
      let y,z;
      assume
B1:   y <==> z;
      assume
B2:   P[y];
      per cases by B1;
      suppose
C1:     y ==> z;
B3:     z is normalizable by A1;
        consider u such that
B4:     u is_normform_of z by B3;
B5:     u is_normform_of y by C1,B4,LemN6;
B6:     x is_normform_of y by A2,B2;
        thus P[z] by B4,B6,B5,A1;
      end;
      suppose
C2:     y <== z;
        thus P[z] by B2,C2,Lem5;
      end;
    end;
A5: for y,z st y <=*=> z & P[y] holds P[z] from Star2(A4);
    thus y =*=> x by A3,A5;
  end;

registration
  cluster WN UN* -> N.F. for ARS;
  coherence by LemN12;

  cluster WN UN* -> UN for ARS;
  coherence;
end;

theorem Lem21:
  y is_normform_of x & z is_normform_of x & y <> z implies x =+=> y
  proof
    assume
A1: y is_normform_of x;
    assume
A2: z is_normform_of x;
    assume
A3: y <> z;
A6: x = y or x =+=> y by A1,LemN;
    thus x =+=> y by A3,A1,A2,A6,LemN1;
  end;

theorem Lem22:
  X is WN UN* implies nf x is_normform_of x
  proof
    assume
A1: X is WN UN*;
A4: x is normalizable by A1;
A3: y is_normform_of x & z is_normform_of x implies y = z by A1;
    thus nf x is_normform_of x by A4,A3,Def17;
  end;

theorem Lem23:
  X is WN UN* & y is_normform_of x implies y = nf x
  proof
    assume
A1: X is WN UN*;
    assume
A2: y is_normform_of x;
A4: for z,u holds z is_normform_of x & u is_normform_of x implies z = u
    by A1;
    thus y = nf x by A2,A4,Def17;
  end;

theorem Lem24:
  X is WN UN* implies nf x is normform
  proof
    assume
A1: X is WN UN*;
A2: nf x is_normform_of x by A1,Lem22;
    thus nf x is normform by A2;
  end;

theorem
  X is WN UN* implies nf nf x = nf x
  proof
    assume
A1: X is WN UN*;
A2: nf x is normform by A1,Lem24;
    thus nf nf x = nf x by A1,A2,LemN2,Lem23;
  end;

theorem Lem26:
  X is WN UN* & x =*=> y implies nf x = nf y
  proof
    assume
A1: X is WN UN*;
    assume
A2: x =*=> y;
A4: nf y is_normform_of x by A2,A1,Lem22,LemN7;
    thus nf x = nf y by A1,A4,Lem23;
  end;

theorem Lem27:
  X is WN UN* & x <=*=> y implies nf x = nf y
  proof
    assume
A1: X is WN UN*;
    assume
A2: x <=*=> y;
    defpred P[Element of X] means nf x = nf $1;
A3: P[x];
A4: for z,u st z <==> u & P[z] holds P[u] by A1,Th2,Lem26;
    P[y] from Star2A(A2,A3,A4);
    hence thesis;
  end;

theorem
  X is WN UN* & nf x = nf y implies x <=*=> y
  proof
    assume
A1: X is WN UN*;
    assume
A2: nf x = nf y;
    nf x is_normform_of x & nf x is_normform_of y by A1,A2,Lem22; then
    x <=*=> nf x & nf x <=*=> y by LemZ;
    hence thesis by Th7;
  end;

begin :: Divergence and Convergence

definition
  let X,x,y;
  pred x <<>> y means
  ex z st x <=*= z & z =*=> y;
  symmetry;
  reflexivity;
  pred x >><< y means:DEF2:
  ex z st x =*=> z & z <=*= y;
  symmetry;
  reflexivity;
  pred x <<01>> y means
  ex z st x <=01= z & z =01=> y;
  symmetry;
  reflexivity;
  pred x >>01<< y means
  ex z st x =01=> z & z <=01= y;
  symmetry;
  reflexivity;
end;

theorem Ch11:
  x <<>> y iff x,y are_divergent_wrt the reduction of X
  proof set R = the reduction of X;
    thus x <<>> y implies x,y are_divergent_wrt R
    proof
      given z such that
A1:   x <=*= z & z =*=> y;
      take z;
      thus R reduces z,x & R reduces z,y by A1;
    end;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    given a being object such that
A2: R reduces a,x & R reduces a,y;
    per cases;
    suppose
      a in field R; then
      reconsider z = a as Element of X by F0;
      take z;
      thus R reduces z,x & R reduces z,y by A2;
    end;
    suppose
      not a in field R; then
      a = x & a = y by A2,REWRITE1:18;
      hence thesis;
    end;
  end;

theorem Ch12:
  x >><< y iff x,y are_convergent_wrt the reduction of X
  proof set R = the reduction of X;
    thus x >><< y implies x,y are_convergent_wrt R
    proof
      given z such that
A1:   z <=*= x & y =*=> z;
      take z;
      thus R reduces x,z & R reduces y,z by A1;
    end;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    given a being object such that
A2: R reduces x,a & R reduces y,a;
    per cases;
    suppose
      a in field R; then
      reconsider z = a as Element of X by F0;
      take z;
      thus R reduces x,z & R reduces y,z by A2;
    end;
    suppose
      not a in field R; then
      a = x & a = y by A2,REWRITE1:18;
      hence thesis;
    end;
  end;

theorem
  x <<01>> y iff x,y are_divergent<=1_wrt the reduction of X
  proof set R = the reduction of X;
    thus x <<01>> y implies x,y are_divergent<=1_wrt R
    proof
      given z such that
A1:   x <=01= z & z =01=> y;
      take z;
      (z ==> x or z = x) & (z ==> y or z = y) by A1;
      hence ([z,x] in R or z = x) & ([z,y] in R or z = y);
    end;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    given a being object such that
A2: ([a,x] in R or a = x) & ([a,y] in R or a = y);
    a in field R or a = x or a = y by A2,RELAT_1:15; then
    reconsider z = a as Element of X by F0;
    take z;
    thus z = x or z ==> x by A2;
    thus z = y or z ==> y by A2;
  end;

theorem Ch14:
  x >>01<< y iff x,y are_convergent<=1_wrt the reduction of X
  proof set R = the reduction of X;
    thus x >>01<< y implies x,y are_convergent<=1_wrt R
    proof
      given z such that
A1:   z <=01= x & y =01=> z;
      take z;
      (x ==> z or z = x) & (y ==> z or z = y) by A1;
      hence ([x,z] in R or x = z) & ([y,z] in R or y = z);
    end;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    given a being object such that
A2: ([x,a] in R or x = a) & ([y,a] in R or y = a);
    a in field R or a = x or a = y by A2,RELAT_1:15; then
    reconsider z = a as Element of X by F0;
    take z;
    thus x = z or x ==> z by A2;
    thus y = z or y ==> z by A2;
  end;

definition
  let X;
  attr X is DIAMOND means
  x <<01>> y implies x >>01<< y;
  attr X is CONF means
  x <<>> y implies x >><< y;
  attr X is CR means
  x <=*=> y implies x >><< y;
  attr X is WCR means
  x <<01>> y implies x >><< y;
end;

definition
  let X;
  attr X is COMP means
  X is SN CONF;
end;

scheme isCR{X() -> non empty ARS, F(Element of X()) -> Element of X()}:
  X() is CR
provided
A1: for x being Element of X() holds x =*=> F(x)
and
A2: for x,y being Element of X() st x <=*=> y holds F(x) = F(y)
  proof
    let x,y be Element of X(); assume x <=*=> y; then
A3: F(x) = F(y) by A2;
    take z = F(x);
    thus thesis by A3,A1;
  end;

Lm3:
  x =*=> y implies x <=*=> y
  proof
    assume
A1: x =*=> y;
    defpred P[Element of X] means x <=*=> $1;
A2: P[x];
A3: for y,z st y ==> z & P[y] holds P[z]
    proof
      let y,z;
      assume
A4:   y ==> z;
      assume
A5:   P[y];
A6:   y <==> z by A4;
A7:   y <=*=> z by A6,Th6;
      thus P[z] by A5,A7,Th7;
    end;
    P[y] from Star1(A1,A2,A3);
    hence thesis;
  end;

Lm2: x <<>> y implies x <=*=> y
  proof
    assume
A1: x <<>> y;
    consider u such that
A2: x <=*= u & u =*=> y by A1;
A3: x <=*=> u & u <=*=> y by A2,Lm3;
    thus x <=*=> y by A3,Th7;
  end;

Lm1: X is CR implies X is CONF by Lm2;

scheme isCOMP{X() -> non empty ARS, F(Element of X()) -> Element of X()}:
  X() is COMP
provided
A1: X() is SN
and
A2: for x being Element of X() holds x =*=> F(x)
and
A3: for x,y being Element of X() st x <=*=> y holds F(x) = F(y)
  proof
    X() is CR from isCR(A2,A3);
    hence X() is SN CONF by A1,Lm1;
  end;

theorem Lem18:
  x <<01>> y implies x <<>> y
  proof
    given z such that
A2: x <=01= z & z =01=> y;
    take z;
    thus x <=*= z & z =*=> y by A2,Lem1;
  end;

theorem Lem18a:
  x >>01<< y implies x >><< y
  proof
    given z such that
A2: x =01=> z & z <=01= y;
    take z;
    thus x =*=> z & z <=*= y by A2,Lem1;
  end;

theorem
  x ==> y implies x <<01>> y
  proof
    assume
A1: x ==> y;
    take x;
    thus x <=01= x & x =01=> y by A1;
  end;

theorem Th17:
  x ==> y implies x >>01<< y
  proof
    assume
A1: x ==> y;
    take y;
    thus x =01=> y & y =01=> y by A1;
  end;

theorem
  x =01=> y implies x <<01>> y;

theorem
  x =01=> y implies x >>01<< y;

theorem
  x <==> y implies x <<01>> y
  proof
    assume
A1: x <==> y;
    per cases by A1;
    suppose
A2:   x ==> y;
      take x;
      thus x <=01= x & x =01=> y by A2;
    end;
    suppose
A3:   x <== y;
      take y;
      thus x <=01= y & y =01=> y by A3;
    end;
  end;

theorem
  x <==> y implies x >>01<< y
  proof
    assume
A1: x <==> y;
    per cases by A1;
    suppose
A2:   x ==> y;
      take y;
      thus x =01=> y & y <=01= y by A2;
    end;
    suppose
A3:   x <== y;
      take x;
      thus x =01=> x & x <=01= y by A3;
    end;
  end;

theorem
  x <=01=> y implies x <<01>> y
  proof
    assume
A1: x <=01=> y;
    per cases by A1,Lem31;
    suppose x = y;
      hence thesis;
    end;
    suppose
A2:   x ==> y;
      take x;
      thus x <=01= x & x =01=> y by A2;
    end;
    suppose
A3:   x <== y;
      take y;
      thus x <=01= y & y =01=> y by A3;
    end;
  end;

theorem
  x <=01=> y implies x >>01<< y
  proof
    assume
A1: x <=01=> y;
    per cases by A1,Lem31;
    suppose x = y;
      hence thesis;
    end;
    suppose
A2:   x ==> y;
      take y;
      thus x =01=> y & y <=01= y by A2;
    end;
    suppose
A3:   x <== y;
      take x;
      thus x =01=> x & x <=01= y by A3;
    end;
  end;

theorem Th17a:
  x ==> y implies x >><< y by Th17,Lem18a;

theorem Lem17:
  x =*=> y implies x >><< y;

theorem
  x =*=> y implies x <<>> y;

theorem
  x =+=> y implies x >><< y
  proof
    assume
A1: x =+=> y;
    take y; thus thesis by A1,Lem2;
  end;

theorem
  x =+=> y implies x <<>> y
  proof
    assume
A1: x =+=> y;
    take x; thus thesis by A1,Lem2;
  end;

theorem Lm11:
  x ==> y & x ==> z implies y <<01>> z
  proof
    assume
A1: x ==> y;
    assume
A2: x ==> z;
    take x;
    thus y <=01= x by A1;
    thus x =01=> z by A2;
  end;

theorem
  x ==> y & z ==> y implies x >>01<< z
  proof
    assume
A1: x ==> y;
    assume
A2: z ==> y;
    take y;
    thus y <=01= x by A1;
    thus z =01=> y by A2;
  end;

theorem
  x >><< z & z <== y implies x >><< y
  proof
    given u such that
A3: x =*=> u & u <=*= z;
    assume
A2: z <== y;
    take u;
    thus x =*=> u by A3;
    thus y =*=> u by A2,A3,Lem5;
  end;

theorem
  x >><< z & z <=01= y implies x >><< y
  proof
    given u such that
A3: x =*=> u & u <=*= z;
    assume
A2: z <=01= y;
    take u;
    thus x =*=> u by A3;
    thus y =*=> u by A2,A3,Lem8;
  end;

theorem Lm5:
  x >><< z & z <=*= y implies x >><< y
  proof
    given u such that
A3: x =*=> u & u <=*= z;
    assume
A2: z <=*= y;
    take u;
    thus x =*=> u by A3;
    thus y =*=> u by A2,A3,Th3;
  end;

theorem Lem19:
  x <<>> y implies x <=*=> y
  proof
    given u such that
A2: x <=*= u & u =*=> y;
A3: x <=*=> u & u <=*=> y by A2,LemZ;
    thus x <=*=> y by A3,Th7;
  end;

theorem
  x >><< y implies x <=*=> y
  proof
    given u such that
A2: x =*=> u & u <=*= y;
A3: x <=*=> u & u <=*=> y by A2,LemZ;
    thus x <=*=> y by A3,Th7;
  end;

begin :: Church-Rosser Property

theorem
  X is DIAMOND iff the reduction of X is subcommutative
  proof
    set R = the reduction of X;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    thus X is DIAMOND implies R is subcommutative
    proof assume
A1:   x <<01>> y implies x >>01<< y;
      let a,b,c be object;
      assume
A2:   [a,b] in R & [a,c] in R; then
      a in field R & b in field R & c in field R by RELAT_1:15; then
      reconsider x = a, y = b, z = c as Element of X by F0;
      x ==> y & x ==> z by A2; then
      x =01=> y & x =01=> z; then
      y <<01>> z;
      hence b,c are_convergent<=1_wrt R by A1,Ch14;
    end;
    assume
A3: for a,b,c being object st [a,b] in R & [a,c] in R
    holds b,c are_convergent<=1_wrt R;
    let x,y; given z such that
A4: x <=01= z & z =01=> y;
    per cases by A4;
    suppose
      x <== z & z ==> y;
      hence thesis by A3,Ch14;
    end;
    suppose
      x = z & z = y;
      hence thesis;
    end;
    suppose
      x <== z & z = y;
      hence thesis by Th17;
    end;
    suppose
      x = z & z ==> y;
      hence thesis by Th17;
    end;
  end;

theorem Ch17:
  X is CONF iff the reduction of X is confluent
  proof
    set R = the reduction of X;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    thus X is CONF implies R is confluent
    proof assume
A1:   x <<>> y implies x >><< y;
      let a,b be object; assume
A2:   a,b are_divergent_wrt R; then
A3:   a,b are_convertible_wrt R by REWRITE1:37;
      per cases by A3,REWRITE1:32;
      suppose
        a in field R & b in field R; then
        reconsider x = a, y = b as Element of X by F0;
        x <<>> y by A2,Ch11;
        hence a,b are_convergent_wrt R by A1,Ch12;
      end;
      suppose
        a = b;
        hence a,b are_convergent_wrt R by REWRITE1:38;
      end;
    end;
    assume
A5: for a,b being object st a,b are_divergent_wrt R
    holds a,b are_convergent_wrt R;
    let x,y; assume x <<>> y; then
    x,y are_divergent_wrt R by Ch11;
    hence thesis by A5,Ch12;
  end;

theorem
  X is CR iff the reduction of X is with_Church-Rosser_property
  proof
    set R = the reduction of X;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    thus X is CR implies R is with_Church-Rosser_property
    proof assume
A1:   x <=*=> y implies x >><< y;
      let a,b be object; assume
A2:   a,b are_convertible_wrt R;
      per cases by A2,REWRITE1:32;
      suppose
        a in field R & b in field R; then
        reconsider x = a, y = b as Element of X by F0;
        x <=*=> y by A2;
        hence a,b are_convergent_wrt R by A1,Ch12;
      end;
      suppose
        a = b;
        hence a,b are_convergent_wrt R by REWRITE1:38;
      end;
    end;
    assume
A5: for a,b being object st a,b are_convertible_wrt R
    holds a,b are_convergent_wrt R;
    let x,y; assume x <=*=> y;
    hence thesis by A5,Ch12;
  end;

theorem
  X is WCR iff the reduction of X is locally-confluent
  proof
    set R = the reduction of X;
    set A = the carrier of X;
F0: field R c= A \/ A by RELSET_1:8;
    thus X is WCR implies R is locally-confluent
    proof assume
A1:   x <<01>> y implies x >><< y;
      let a,b,c be object; assume
A2:   [a,b] in R & [a,c] in R; then
      a in field R & b in field R & c in field R by RELAT_1:15; then
      reconsider x = a, y = b, z = c as Element of X by F0;
      x ==> y & x ==> z by A2; then
      x =01=> y & x =01=> z; then
      y <<01>> z;
      hence b,c are_convergent_wrt R by A1,Ch12;
    end;
    assume
A3: for a,b,c being object st [a,b] in R & [a,c] in R
    holds b,c are_convergent_wrt R;
    let x,y; given z such that
A4: x <=01= z & z =01=> y;
    per cases by A4;
    suppose
      x <== z & z ==> y;
      hence thesis by A3,Ch12;
    end;
    suppose
      x = z & z = y;
      hence thesis;
    end;
    suppose
      x <== z & z = y;
      hence thesis by Th17a;
    end;
    suppose
      x = z & z ==> y;
      hence thesis by Th17a;
    end;
  end;

theorem
  for X being non empty ARS holds X is COMP iff the reduction of X is complete
  proof let X be non empty ARS;
    set R = the reduction of X;
A2: X is CONF iff R is confluent by Ch17;
    X is SN iff R is strongly-normalizing by Ch7,Ch8;
    hence thesis by A2;
  end;

theorem LemA:
  X is DIAMOND & x <=*= z & z =01=> y implies
  ex u st x =01=> u & u <=*= y
  proof
    assume
A1: for x,y st x <<01>> y holds x >>01<< y;
    assume
A2: x <=*= z;
    assume
A3: z =01=> y;
    defpred P[Element of X] means ex u st $1 =01=> u & u <=*= y;
A4: for u,v st u ==> v & P[u] holds P[v]
    proof
      let u,v;
      assume u ==> v; then
B1:   u =01=> v;
      given w such that
B2:   u =01=> w & w <=*= y;
      v <<01>> w by B1,B2; then
      v >>01<< w by A1; then
      consider u such that
B3:   v =01=> u & u <=01= w;
      thus P[v] by B2,B3,Lem11;
    end;
A5: for u,v st u =*=> v & P[u] holds P[v] from Star(A4);
    thus thesis by A5,A2,A3;
  end;

theorem
  X is DIAMOND & x <=01= y & y =*=> z implies
  ex u st x =*=> u & u <=01= z
  proof
    assume X is DIAMOND & x <=01= y & y =*=> z; then
    ex u st z =01=> u & u <=*= x by LemA;
    hence thesis;
  end;

registration
  cluster DIAMOND -> CONF for ARS;
  coherence
  proof let X;
    assume
A1: X is DIAMOND;
    let x,y;
    given z such that
A2: x <=*= z and
A3: z =*=> y;
    defpred P[Element of X] means x >><< $1;
A4: P[z] by A2,Lem17;
A5: for u,v st u ==> v & P[u] holds P[v]
    proof
      let u,v;
      assume
A6:   u ==> v;
      given w such that
A7:   x =*=> w & w <=*= u;
A8:   u =01=> v by A6;
      consider a such that
A9:   w =01=> a & a <=*= v by A1,A7,A8,LemA;
A10:  x =*=> a by A7,A9,Lem11;
      thus P[v] by A9,A10,DEF2;
    end;
    P[y] from Star1(A3,A4,A5);
    hence x >><< y;
  end;
end;

registration
  cluster DIAMOND -> CR for ARS;
  coherence
  proof let X;
    assume
A1: X is DIAMOND;
    let x,y;
    assume
A2: x <=*=> y;
    defpred P[Element of X] means x >><< $1;
A4: P[x];
A5: for u,v st u <==> v & P[u] holds P[v]
    proof
      let u,v;
      assume
A6:   u <==> v;
      given w such that
A7:   x =*=> w & w <=*= u;
      per cases by A6;
      suppose u ==> v; then
A8:     u =01=> v;
        consider a such that
A9:     w =01=> a & a <=*= v by A1,A7,A8,LemA;
A10:    x =*=> a by A7,A9,Lem11;
        thus P[v] by A9,A10,DEF2;
      end;
      suppose u <== v; then
A11:    v =*=> w by A7,Lem5;
        thus P[v] by A7,A11,DEF2;
      end;
    end;
    P[y] from Star2A(A2,A4,A5);
    hence x >><< y;
  end;
end;

registration
  cluster CR -> WCR for ARS;
  coherence
  proof let X;
    assume
A1: X is CR;
    let x,y;
    assume
A2: x <<01>> y;
A4: x <=*=> y by A2,Lem18,Lem19;
    thus x >><< y by A1,A4;
  end;
end;

registration
  cluster CR -> CONF for ARS;
  coherence by Lm1;
end;

registration
  cluster CONF -> CR for ARS;
  coherence
  proof let X;
    assume
A1: X is CONF;
    let x;
    defpred P[Element of X] means x >><< $1;
A3: for y,z st y <==> z & P[y] holds P[z]
    proof
      let y,z;
      assume
B1:   y <==> z & P[y];
      consider u such that
B2:   x =*=> u & u <=*= y by B1,DEF2;
      per cases by B1;
      suppose
B3:     y ==> z;
        y =*=> z by B3,Th2; then
        u <<>> z by B2;
        hence P[z] by A1,B2,Lm5;
      end;
      suppose
B5:     y <== z;
        thus P[z] by B1,B5,Th2,Lm5;
      end;
    end;
    for y,z st y <=*=> z & P[y] holds P[z] from Star2(A3);
    hence thesis;
  end;
end;

theorem
  X is non CONF WN implies
  ex x,y,z st y is_normform_of x & z is_normform_of x & y <> z
  proof
    given a,b such that
A1: a <<>> b  & not a >><< b;
    consider x such that
A0: a <=*= x & x =*=> b by A1;
    assume
A2: c is normalizable; then
    a is normalizable; then
    consider y such that
A3: y is_normform_of a;
    b is normalizable by A2; then
    consider z such that
A4: z is_normform_of b;
    take x,y,z;
    thus y is_normform_of x & z is_normform_of x by A0,A3,A4,LemN7;
    thus thesis by A1,A3,A4;
  end;

registration
::$N Newman's lemma
  cluster SN WCR -> CR for ARS;
  coherence
  proof let X;
    assume
A1: X is SN WCR;
    assume
A2: X is not CR;
A3: X is not CONF by A2;
    consider x1,x2 being Element of X such that
A4: x1 <<>> x2 & not x1 >><< x2 by A3;
    defpred P[Element of X] means
    ex x,y st x is_normform_of $1 & y is_normform_of $1 & x <> y;
A5: ex x st P[x]
    proof
      consider x such that
B1:   x1 <=*= x & x =*=> x2 by A4;
      take x;
      consider y1 being Element of X such that
B2:   y1 is_normform_of x1 by A1,ThWN1;
      consider y2 being Element of X such that
B3:   y2 is_normform_of x2 by A1,ThWN1;
      take y1,y2;
      thus y1 is_normform_of x by B1,B2,LemN7;
      thus y2 is_normform_of x by B1,B3,LemN7;
      assume
B4:   y1 = y2;
      thus contradiction by A4,B2,B3,B4;
    end;
A6: for x st P[x] ex y st P[y] & x ==> y
    proof
      let x;
      assume P[x]; then
      consider x1,x2 being Element of X such that
C1:   x1 is_normform_of x & x2 is_normform_of x & x1 <> x2;
      x =+=> x1 by C1,Lem21; then
      consider y1 being Element of X such that
C2:   x ==> y1 & y1 =*=> x1;
      x =+=> x2 by C1,Lem21; then
      consider y2 being Element of X such that
C3:   x ==> y2 & y2 =*=> x2;
      y1 >><< y2 by A1,C2,C3,Lm11; then
      consider y such that
C4:   y1 =*=> y & y <=*= y2;
      consider y0 being Element of X such that
C5:   y0 is_normform_of y by A1,ThWN1;
      per cases;
      suppose
D1:     y0 = x1;
        take y2;
D2:     y0 is_normform_of y2 by C4,C5,LemN7;
        x2 is_normform_of y2 by C1,C3;
        hence P[y2] by C1,D1,D2;
        thus x ==> y2 by C3;
      end;
      suppose
D3:     y0 <> x1;
        take y1;
D4:     y0 is_normform_of y1 by C4,C5,LemN7;
        x1 is_normform_of y1 by C1,C2;
        hence thesis by C2,D3,D4;
      end;
    end;
A7: X is not SN from notSN(A5,A6);
    thus contradiction by A1,A7;
  end;
end;

registration
  cluster CR -> N.F. for ARS;
  coherence
  proof let X;
    assume
A1: X is CR;
    let x,y;
    assume
A2: x is normform;
    assume
A3: x <=*=> y;
A4: x >><< y by A1,A3;
    consider z such that
A5: x =*=> z & z <=*= y by A4;
    thus y =*=> x by A2,A5,LemN1;
  end;
end;

registration
  cluster WN UN -> CR for ARS;
  coherence
  proof let X;
    assume
A1: X is WN;
    assume
A2: X is UN;
    let x,y;
    assume
A3: x <=*=> y;
A4: x is normalizable & y is normalizable by A1;
    consider u such that
A5: u is_normform_of x by A4;
    consider v such that
A6: v is_normform_of y by A4;
A7: u is normform & x =*=> u by A5;
    take u;
    thus x =*=> u by A5;
    u <=*=> x by A5,LemZ; then
    u <=*=> y & y <=*=> v by A3,A6,Th7,LemZ;
    hence y =*=> u by A2,A7,A6,Th7;
  end;
end;

registration
  cluster SN CR -> COMP for ARS;
  coherence;

  cluster COMP -> CR WCR N.F. UN UN* WN for ARS;
  coherence;
end;

theorem
  X is COMP implies
  for x,y st x <=*=> y holds nf x = nf y by Lem27;

registration
  cluster WN UN* -> CR for ARS;
  coherence;

  cluster SN UN* -> COMP for ARS;
  coherence;
end;

begin :: Term Rewriting Systems

definition
  struct(ARS,UAStr) TRSStr (#
    carrier -> set,
    charact -> PFuncFinSequence of the carrier,
    reduction -> Relation of the carrier
  #);
end;

registration
  cluster non empty non-empty strict for TRSStr;
  existence
  proof
    set S = the non empty set;
    set o = the non-empty non empty PFuncFinSequence of S;
    set r = the Relation of S;
    take X = TRSStr(#S, o, r#);
    thus the carrier of X is non empty;
    thus the charact of X <> {};
    thus thesis;
  end;
end;

definition
  let S be non empty UAStr;
  attr S is Group-like means
  Seg 3 c= dom the charact of S &
  for f being non empty homogeneous
    PartFunc of (the carrier of S)*, the carrier of S holds
  (f = (the charact of S).1 implies arity f = 0) &
  (f = (the charact of S).2 implies arity f = 1) &
  (f = (the charact of S).3 implies arity f = 2);
end;

theorem Th01:
  for X being non empty set
  for f1,f2,f3 being non empty homogeneous PartFunc of X*, X
   st arity f1 = 0 & arity f2 = 1 & arity f3 = 2
  for S being non empty UAStr
   st the carrier of S = X & <*f1,f2,f3*> c= the charact of S
  holds S is Group-like
  proof
    let X be non empty set;
    let f1,f2,f3 be non empty homogeneous PartFunc of X*, X;
    assume
01: arity f1 = 0;
    assume
02: arity f2 = 1;
    assume
03: arity f3 = 2;
    let S be non empty UAStr;
    assume
04: the carrier of S = X & <*f1,f2,f3*> c= the charact of S;
05: dom <*f1,f2,f3*> = Seg 3 by FINSEQ_2:124;
    hence Seg 3 c= dom the charact of S by 04,RELAT_1:11;
    let f be non empty homogeneous
    PartFunc of (the carrier of S)*, the carrier of S;
    1 in Seg 3 & 2 in Seg 3 & 3 in Seg 3 by FINSEQ_3:1,ENUMSET1:def 1; then
    (the charact of S).1 = <*f1,f2,f3*>.1 &
    (the charact of S).2 = <*f1,f2,f3*>.2 &
    (the charact of S).3 = <*f1,f2,f3*>.3 by 04,05,GRFUNC_1:2;
    hence (f = (the charact of S).1 implies arity f = 0) &
    (f = (the charact of S).2 implies arity f = 1) &
    (f = (the charact of S).3 implies arity f = 2) by 01,02,03,FINSEQ_1:45;
  end;

theorem Th02:
  for X being non empty set
  for f1,f2,f3 being non empty quasi_total homogeneous PartFunc of X*, X
  for S being non empty UAStr
   st the carrier of S = X & <*f1,f2,f3*> = the charact of S
  holds S is quasi_total partial
  proof
    let X be non empty set;
    let f1,f2,f3 be non empty quasi_total homogeneous PartFunc of X*, X;
    let S be non empty UAStr;
    assume
04: the carrier of S = X & <*f1,f2,f3*> = the charact of S;
    set A = the carrier of S;
    thus S is quasi_total
    proof
      let i be Nat, h being PartFunc of A*,A;
      assume i in dom the charact of S; then
      i in Seg 3 by 04,FINSEQ_1:89; then
      i = 1 or i = 2 or i = 3 by FINSEQ_3:1,ENUMSET1:def 1;
      hence thesis by 04,FINSEQ_1:45;
    end;
    let i be Nat, h being PartFunc of A*,A;
    assume i in dom the charact of S; then
    i in Seg 3 by 04,FINSEQ_1:89; then
    i = 1 or i = 2 or i = 3 by FINSEQ_3:1,ENUMSET1:def 1;
    hence thesis by 04,FINSEQ_1:45;
  end;

definition
  let S be non empty non-empty UAStr;
  let o be operation of S;
  let a be Element of dom o;
  redefine func o.a -> Element of S;
  coherence
  proof
    o in rng the charact of S; then
    o <> {} & o in PFuncs((the carrier of S)*, the carrier of S)
    by RELAT_1:def 9; then
    o.a in rng o & rng o c= the carrier of S by RELAT_1:def 19,FUNCT_1:3;
    hence thesis;
  end;
end;

registration
  let S be non empty non-empty UAStr;
  cluster -> non empty for operation of S;
  coherence by RELAT_1:def 9;
  let o be operation of S;
  cluster -> Relation-like Function-like for Element of dom o;
  coherence
  proof
    let a be Element of dom o;
    a in dom o & dom o c= (the carrier of S)*; then
    a is Element of (the carrier of S)*;
    hence thesis;
  end;
end;

registration
  let S be partial non empty non-empty UAStr;
  cluster -> homogeneous for operation of S;
  coherence
  proof
    let o be operation of S;
    consider i being object such that
A1: i in dom the charact of S & o = (the charact of S).i by FUNCT_1:def 3;
    thus thesis by A1;
  end;
end;

registration
  let S be quasi_total non empty non-empty UAStr;
  cluster -> quasi_total for operation of S;
  coherence
  proof
    let o be operation of S;
    consider i being object such that
A1: i in dom the charact of S & o = (the charact of S).i by FUNCT_1:def 3;
    thus thesis by A1,MARGREL1:def 24;
  end;
end;

theorem ThA:
  for S being non empty non-empty UAStr st S is Group-like
  holds
    1 is OperSymbol of S & 2 is OperSymbol of S & 3 is OperSymbol of S
    proof
      let S be non empty non-empty UAStr;
      assume
A0:   Seg 3 c= dom the charact of S;
      1 in Seg 3 & 2 in Seg 3 & 3 in Seg 3 by FINSEQ_3:1,ENUMSET1:def 1;
      hence thesis by A0;
    end;

theorem ThB:
  for S being partial non empty non-empty UAStr st S is Group-like
  holds
    arity Den(In(1, dom the charact of S), S) = 0 &
    arity Den(In(2, dom the charact of S), S) = 1 &
    arity Den(In(3, dom the charact of S), S) = 2
  proof
    let S be partial non empty non-empty UAStr;
    assume
A1: S is Group-like; then
    1 is OperSymbol of S & 2 is OperSymbol of S & 3 is OperSymbol of S
    by ThA; then
    In(1, dom the charact of S) = 1 &
    In(2, dom the charact of S) = 2 &
    In(3, dom the charact of S) = 3;
    hence thesis by A1,PUA2MSS1:def 1;
  end;

definition
  let S be non empty non-empty TRSStr;
  attr S is invariant means:
DEF2:
  for o being OperSymbol of S
  for a,b being Element of dom Den(o,S)
  for i being Nat st i in dom a
  for x,y being Element of S
  st x = a.i & b = a+*(i,y) & x ==> y
  holds Den(o,S).a ==> Den(o,S).b;
end;

definition
  let S be non empty non-empty TRSStr;
  attr S is compatible means
  for o being OperSymbol of S
  for a,b being Element of dom Den(o,S)
  st for i being Nat st i in dom a holds
     for x,y being Element of S st x = a.i & y = b.i holds x ==> y
  holds Den(o,S).a =*=> Den(o,S).b;
end;

theorem Th0:
  for n being natural number, X being non empty set, x being Element of X
  ex f being non empty homogeneous quasi_total PartFunc of X*, X st
  arity f = n & f = (n-tuples_on X) --> x
  proof
    let n be natural number, X be non empty set;
    let x be Element of X;
    set f = (n-tuples_on X) --> x;
A1: dom f = n-tuples_on X & rng f = {x} & n in omega
    by FUNCOP_1:8,ORDINAL1:def 12; then
    dom f c= X* & rng f c= X by ZFMISC_1:31,FINSEQ_2:134; then
    reconsider f as non empty PartFunc of X*, X by RELSET_1:4;
A2: f is quasi_total
    proof
      let x,y be FinSequence of X; assume
      len x = len y & x in dom f; then
      len x = n & len y = n by A1,FINSEQ_2:132;
      hence thesis by FINSEQ_2:133;
    end;
    f is homogeneous
    proof
      let x,y be FinSequence; assume
      x in dom f & y in dom f; then
      reconsider x,y as Element of n-tuples_on X;
      len x = n & len y = n by A1,FINSEQ_2:132;
      hence thesis;
    end; then
    reconsider f as non empty homogeneous quasi_total PartFunc of X*, X by A2;
    take f;
    set y = the Element of n-tuples_on X;
A3: for x being FinSequence st x in dom f holds n = len x by A1,FINSEQ_2:132;
    y in dom f;
    hence arity f = n by A3,MARGREL1:def 25;
    thus thesis;
  end;

registration
  let X be non empty set;
  let O be PFuncFinSequence of X;
  let r be Relation of X;
  cluster TRSStr(#X, O, r#) -> non empty;
  coherence;
end;

registration
  let X be non empty set;
  let O be non empty non-empty PFuncFinSequence of X;
  let r be Relation of X;
  cluster TRSStr(#X, O, r#) -> non-empty;
  coherence
  proof
    thus the charact of TRSStr(#X, O, r#) <> {};
    thus the charact of TRSStr(#X, O, r#) is non-empty;
  end;
end;

definition
  let X be non empty set;
  let x be Element of X;
  func TotalTRS(X,x) -> non empty non-empty strict TRSStr means:
DEF3:
  the carrier of it = X &
  the charact of it =
  <*(0-tuples_on X)-->x, (1-tuples_on X)-->x, (2-tuples_on X)-->x*> &
  the reduction of it = nabla X;
  uniqueness;
  existence
  proof
    consider f0 being non empty homogeneous quasi_total PartFunc of X*, X
    such that
A0: arity f0 = 0 & f0 = (0-tuples_on X) --> x by Th0;
    consider f1 being non empty homogeneous quasi_total PartFunc of X*, X
    such that
A1: arity f1 = 1 & f1 = (1-tuples_on X) --> x by Th0;
    consider f2 being non empty homogeneous quasi_total PartFunc of X*, X
    such that
A2: arity f2 = 2 & f2 = (2-tuples_on X) --> x by Th0;
    set r = nabla X;
    reconsider a = f0, b = f1, c = f2 as Element of PFuncs(X*, X)
    by PARTFUN1:45;
    reconsider O = <*a,b,c*> as non empty non-empty PFuncFinSequence of X;
    take S = TRSStr(#X, O, r#);
    thus thesis by A0,A1,A2;
  end;
end;

registration
  let X be non empty set;
  let x be Element of X;
  cluster TotalTRS(X,x) -> quasi_total partial Group-like invariant;
  coherence
  proof set S = TotalTRS(X,x);
A3: the carrier of S = X & the charact of S =
    <*(0-tuples_on X)-->x, (1-tuples_on X)-->x, (2-tuples_on X)-->x*> &
    the reduction of S = nabla X by DEF3;
    consider f0 being non empty homogeneous quasi_total PartFunc of X*, X
    such that
A0: arity f0 = 0 & f0 = (0-tuples_on X) --> x by Th0;
    consider f1 being non empty homogeneous quasi_total PartFunc of X*, X
    such that
A1: arity f1 = 1 & f1 = (1-tuples_on X) --> x by Th0;
    consider f2 being non empty homogeneous quasi_total PartFunc of X*, X
    such that
A2: arity f2 = 2 & f2 = (2-tuples_on X) --> x by Th0;
    [:X,X:] c= [:X,X:]; then
    reconsider r = [:X,X:] as Relation of X;
    reconsider a = f0, b = f1, c = f2 as Element of PFuncs(X*, X)
    by PARTFUN1:45;
    thus S is quasi_total partial Group-like by A0,A1,A2,A3,Th01,Th02;
    let o be OperSymbol of S;
    let a,b be Element of dom Den(o,S);
    let i be Nat such that i in dom a;
    let x,y be Element of S such that x = a.i & b = a+*(i,y) & x ==> y;
    thus [Den(o,S).a,Den(o,S).b] in the reduction of S by A3,ZFMISC_1:87;
  end;
end;

registration
  cluster strict quasi_total partial Group-like invariant for
  non empty non-empty TRSStr;
  existence
  proof
    take TotalTRS(NAT,In(0,NAT));
    thus thesis;
  end;
end;

definition
  let S be Group-like quasi_total partial non empty non-empty TRSStr;
  func 1.S -> Element of S equals
     Den(In(1,dom the charact of S), S).{};
  coherence
  proof
    arity Den(In(1,dom the charact of S), S) = 0 by ThB; then
    dom Den(In(1,dom the charact of S), S) = 0-tuples_on the carrier of S
    by COMPUT_1:22 .= {{}} by COMPUT_1:5; then
    {} in dom Den(In(1,dom the charact of S), S) by TARSKI:def 1;
    hence thesis by FUNCT_1:102;
  end;
  let a be Element of S;
  func a " -> Element of S equals
     Den(In(2,dom the charact of S), S).<*a*>;
  coherence
  proof
    arity Den(In(2,dom the charact of S), S) = 1 by ThB; then
    dom Den(In(2,dom the charact of S), S) = 1-tuples_on the carrier of S &
    <*a*> is Element of 1-tuples_on the carrier of S
    by FINSEQ_2:98,MARGREL1:22;
    hence thesis by FUNCT_1:102;
  end;
  let b be Element of S;
  func a * b -> Element of S equals
     Den(In(3,dom the charact of S), S).<*a,b*>;
  coherence
  proof
    arity Den(In(3,dom the charact of S), S) = 2 by ThB; then
    dom Den(In(3,dom the charact of S), S) = 2-tuples_on the carrier of S &
    <*a,b*> is Element of 2-tuples_on the carrier of S
    by FINSEQ_2:101,MARGREL1:22;
    hence thesis by FUNCT_1:102;
  end;
end;

reserve
  S for Group-like quasi_total partial invariant non empty non-empty TRSStr;
reserve a,b,c for Element of S;

theorem
  a ==> b implies a" ==> b"
  proof
    assume
A0: a ==> b;
    set o = In(2, dom the charact of S);
    arity Den(o, S) = 1 by ThB; then
    dom Den(o, S) = 1-tuples_on the carrier of S by MARGREL1:22; then
    reconsider aa = <*a*>, bb = <*b*> as Element of dom Den(o, S)
    by FINSEQ_2:98;
A2: dom <*a*> = Seg 1 & 1 in Seg 1 by FINSEQ_1:1,38;
A3: <*a*>.1 = a by FINSEQ_1:40;
    <*a*>+*(1,b) = <*b*> by FUNCT_7:95; then
    Den(o,S).aa ==> Den(o,S).bb by A0,A2,A3,DEF2;
    hence a" ==> b";
  end;

theorem ThI2:
  a ==> b implies a*c ==> b*c
  proof
    assume
A0: a ==> b;
    set o = In(3, dom the charact of S);
    arity Den(o, S) = 2 by ThB; then
    dom Den(o, S) = 2-tuples_on the carrier of S by MARGREL1:22; then
    reconsider ac = <*a,c*>, bc = <*b,c*> as Element of dom Den(o, S)
    by FINSEQ_2:101;
A2: dom <*a,c*> = Seg 2 & 1 in Seg 2 by FINSEQ_1:1,89;
A3: <*a,c*>.1 = a by FINSEQ_1:44;
    <*a,c*>+*(1,b) = <*b,c*> by COMPUT_1:1; then
    Den(o,S).ac ==> Den(o,S).bc by A0,A2,A3,DEF2;
    hence a*c ==> b*c;
  end;

theorem ThI3:
  a ==> b implies c*a ==> c*b
  proof
    assume
A0: a ==> b;
    set o = In(3, dom the charact of S);
    arity Den(o, S) = 2 by ThB; then
    dom Den(o, S) = 2-tuples_on the carrier of S by MARGREL1:22; then
    reconsider ac = <*c,a*>, bc = <*c,b*> as Element of dom Den(o, S)
    by FINSEQ_2:101;
A2: dom <*c,a*> = Seg 2 & 2 in Seg 2 by FINSEQ_1:1,89;
A3: <*c,a*>.2 = a by FINSEQ_1:44;
    <*c,a*>+*(2,b) = <*c,b*> by COMPUT_1:1; then
    Den(o,S).ac ==> Den(o,S).bc by A0,A2,A3,DEF2;
    hence c*a ==> c*b;
  end;

begin :: An Execution of Knuth-Bendix Algorithm

reserve S for Group-like quasi_total partial non empty non-empty TRSStr;
reserve a,b,c for Element of S;

definition
  let S;
  attr S is (R1) means
   1.S * a ==> a;
  attr S is (R2) means
   a" * a ==> 1.S;
  attr S is (R3) means
   (a * b) * c ==> a * (b * c);
  attr S is (R4) means
   a" * (a * b) ==> b;
  attr S is (R5) means
   (1.S)" * a ==> a;
  attr S is (R6) means
   (a")" * 1.S ==> a;
  attr S is (R7) means
   (a")" * b ==> a * b;
  attr S is (R8) means
   a * 1.S ==> a;
  attr S is (R9) means
   (a")" ==> a;
  attr S is (R10) means
   (1.S)" ==> 1.S;
  attr S is (R11) means
   a * (a") ==> 1.S;
  attr S is (R12) means
   a * (a" * b) ==> b;
  attr S is (R13) means
   a * (b * (a * b)") ==> 1.S;
  attr S is (R14) means
   a * (b * a)" ==> b";
  attr S is (R15) means
   (a * b)" ==> b" * a";
end;

reserve
  S for Group-like quasi_total partial invariant non empty non-empty TRSStr,
  a,b,c for Element of S;

theorem
  S is (R1) (R2) (R3) implies a" * (a * b) <<>> b
  proof
    assume
A1: S is (R1) (R2) (R3);
    take (a"*a)*b;
    thus (a"*a)*b =*=> a"*(a*b) by A1,Th2;
    (a"*a)*b ==> 1.S * b & 1.S * b ==> b by A1,ThI2; then
    (a"*a)*b =*=> 1.S * b & 1.S * b =*=> b by Th2;
    hence (a"*a)*b =*=> b by Th3;
  end;

theorem
  S is (R1) (R4) implies (1.S)" * a <<>> a
  proof
    assume
A1: S is (R1) (R4);
    take (1.S)"*(1.S*a);
    1.S*a ==> a by A1;
    hence (1.S)"*(1.S*a) =*=> (1.S)" * a by Th2,ThI3;
    thus thesis by A1,Th2;
  end;

theorem
  S is (R2) (R4) implies (a")" * 1.S <<>> a
  proof
    assume
A1: S is (R2) (R4);
    take (a")" * (a" * a);
    a" * a ==> 1.S by A1;
    hence (a")" * (a" * a) =*=> (a")" * 1.S by Th2,ThI3;
    thus (a")" * (a" * a) =*=> a by A1,Th2;
  end;

theorem
  S is (R1) (R3) (R6) implies (a")" * b <<>> a * b
  proof
    assume
A1: S is (R1) (R3) (R6);
    take (a""*1.S)*b;
A2: (a""*1.S)*b =*=> a""*(1.S*b) by A1,Th2;
    1.S*b ==> b by A1; then
    a""*(1.S*b) =*=> a""*b by Th2,ThI3;
    hence (a""*1.S)*b =*=> a""*b by A2,Th3;
    a"" * 1.S ==> a by A1;
    hence (a"" * 1.S) * b =*=> a * b by Th2,ThI2;
  end;

theorem
  S is (R6) (R7) implies a * 1.S <<>> a
  proof
    assume
A1: S is (R6) (R7);
    take a""*1.S;
    thus a""*1.S =*=> a*1.S by A1,Th2;
    thus a"" * 1.S =*=> a by A1,Th2;
  end;

theorem
  S is (R6) (R8) implies (a")" <<>> a
  proof
    assume
A1: S is (R6) (R8);
    take a""*1.S;
    thus a""*1.S =*=> a"" by A1,Th2;
    thus a"" * 1.S =*=> a by A1,Th2;
  end;

theorem
  S is (R5) (R8) implies (1.S)" <<>> 1.S
  proof
    assume
A1: S is (R5) (R8);
    take (1.S)"*1.S;
    thus (1.S)"*1.S =*=> (1.S)" by A1,Th2;
    thus (1.S)" * 1.S =*=> 1.S by A1,Th2;
  end;

theorem
  S is (R2) (R9) implies a * (a") <<>> 1.S
  proof
    assume
A1: S is (R2) (R9);
    take a""*a";
    a"" ==> a by A1;
    hence a""*a" =*=> a*a" by Th2,ThI2;
    thus a""*a" =*=> 1.S by A1,Th2;
  end;

theorem
  S is (R1) (R3) (R11) implies a * (a" * b) <<>> b
  proof
    assume
A1: S is (R1) (R3) (R11);
    take (a * a") * b;
    thus (a * a") * b =*=> a * (a" * b) by A1,Th2;
    (a * a") * b ==> 1.S * b & 1.S * b ==> b by A1,ThI2;
    hence (a * a") * b =*=> b by Lem3;
  end;

theorem
  S is (R3) (R11) implies a * (b * (a * b)") <<>> 1.S
  proof
    assume
A1: S is (R3) (R11);
    take (a * b) * (a * b)";
    thus (a * b) * (a * b)" =*=> a * (b * (a * b)") by A1,Th2;
    thus (a * b) * (a * b)" =*=> 1.S by A1,Th2;
  end;

theorem
  S is (R4) (R8) (R13) implies a * (b * a)" <<>> b"
  proof
    assume
A1: S is (R4) (R8) (R13);
    take b"*(b*(a*(b*a)"));
    thus b"*(b*(a*(b*a)")) =*=> a*(b*a)" by A1,Th2;
    b"*(b*(a*(b*a)")) ==> b"*1.S & b"*1.S ==> b" by A1,ThI3;
    hence b"*(b*(a*(b*a)")) =*=> b" by Lem3;
  end;

theorem
  S is (R4) (R14) implies (a * b)" <<>> b" * a"
  proof
    assume
A1: S is (R4) (R14);
    take b"*(b*(a*b)");
    thus b"*(b*(a*b)") =*=> (a * b)" by A1,Th2;
    (b*(a*b)") ==> a" by A1;
    hence b"*(b*(a*b)") =*=> b" * a" by Th2,ThI3;
  end;

theorem
  S is (R1) (R10) implies (1.S)" * a =*=> a
  proof
    assume
A1: S is (R1) (R10);
    (1.S)"*a ==> 1.S*a & 1.S*a ==> a by A1,ThI2;
    hence (1.S)" * a =*=> a by Lem3;
  end;

theorem
  S is (R8) (R9) implies (a")" * 1.S =*=> a
  proof
    assume S is (R8) (R9); then
    (a")" * 1.S ==> a"" & a"" ==> a;
    hence (a")" * 1.S =*=> a by Lem3;
  end;

theorem
  S is (R9) implies (a")" * b =*=> a * b
  proof
    assume S is (R9); then
    a"" ==> a;
    hence (a")" * b =*=> a * b by Th2,ThI2;
  end;

theorem
  S is (R11) (R14) implies a * (b * (a * b)") =*=> 1.S
  proof
    assume
A1: S is (R11) (R14);
    a * (b * (a * b)") ==> a*a" & a*a" ==> 1.S by A1,ThI3;
    hence a * (b * (a * b)") =*=> 1.S by Lem3;
  end;

theorem
  S is (R12) (R15) implies a * (b * a)" =*=> b"
  proof
    assume
A1: S is (R12) (R15);
    a * (b * a)" ==> a*(a"*b") & a*(a"*b") ==> b" by A1,ThI3;
    hence a * (b * a)" =*=> b" by Lem3;
  end;