:: Abian's Fixed Point Theorem
::  by Piotr Rudnicki and Andrzej Trybulec

environ

 vocabularies NUMBERS, SETFAM_1, FUNCT_1, SUBSET_1, INT_1, RELAT_1, CARD_1,
      XXREAL_0, ARYTM_3, ARYTM_1, FUNCT_7, XBOOLE_0, TARSKI, ZFMISC_1,
      FINSET_1, EQREL_1, FUNCOP_1, ABIAN, XCMPLX_0, NAT_1, RECDEF_2;
 notations TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0,
      MCART_1, DOMAIN_1, FINSET_1, SETFAM_1, SEQ_4, RELAT_1, FUNCT_1, FUNCT_2,
      FUNCOP_1, INT_1, NAT_1, NAT_D, EQREL_1, FUNCT_7, XXREAL_0;
 constructors SETFAM_1, XXREAL_0, REAL_1, NAT_1, NAT_D, EQREL_1, SEQ_4,
      REALSET1, FUNCT_7, XXREAL_2, RELSET_1;
 registrations XBOOLE_0, SUBSET_1, SETFAM_1, FUNCT_1, ORDINAL1, RELSET_1,
      PARTFUN1, FINSET_1, XREAL_0, INT_1, MEMBERED, EQREL_1, XXREAL_2,
      XXREAL_0, NAT_1;
 requirements REAL, NUMERALS, BOOLE, SUBSET, ARITHM;
 definitions TARSKI, XBOOLE_0, RELAT_1, FUNCT_1, SETFAM_1, INT_1;
 equalities RELAT_1;
 expansions XBOOLE_0, INT_1;
 theorems TARSKI, ZFMISC_1, FUNCT_1, FUNCT_2, EQREL_1, NAT_1, INT_1, MCART_1,
      SCHEME1, FUNCOP_1, SETFAM_1, XBOOLE_0, XBOOLE_1, FUNCT_7, XREAL_1,
      XXREAL_0, NAT_D, XXREAL_2;
 schemes EQREL_1, FUNCT_2, NAT_1, DOMAIN_1;

begin

reserve x, y, z, E, E1, E2, E3 for set,
  sE for Subset-Family of E,
  f for Function of E, E,
  k, l, m, n for Nat;

definition
  let i be Integer;
  attr i is even means
  2 divides i;
end;

notation
  let i be Integer;
  antonym i is odd for i is even;
end;

Lm1:
  for i being Integer holds i is even iff ex j being Integer st i = 2*j
    by INT_1:def 3;

definition
  let n be Nat;
  redefine attr n is even means
  ex k st n = 2*k;
  compatibility
  proof
    hereby
      assume n is even;
      then 2 divides n;
      then consider k being Integer such that
A1:   n = 2*k;
      0<=k by A1,XREAL_1:132;
      then k in NAT by INT_1:3;
      then reconsider k as Nat;
      take k;
      thus n = 2*k by A1;
    end;
    thus thesis by Lm1;
  end;
end;

registration
  cluster even for Nat;
  existence
  proof
    take 0, 0;
    thus thesis;
  end;
  cluster odd for Nat;
  existence
  proof
    take 1;
    let k be Nat;
    thus thesis by NAT_1:15;
  end;
  cluster even for Element of NAT;
  existence
  proof
    take 0, 0;
    thus thesis;
  end;
  cluster odd for Element of NAT;
  existence
  proof
    take 1;
    let k be Nat;
    thus thesis by NAT_1:15;
  end;
  cluster even for Integer;
  existence
  proof
    take 0, 0;
    thus thesis;
  end;
  cluster odd for Integer;
  existence
  proof
    take 1;
    assume 1 is even;
    then ex k being Integer st 1 = 2*k;
    hence contradiction by INT_1:9;
  end;
end;

theorem Th1:
  for i being Integer holds i is odd iff ex j being Integer st i = 2*j+1
proof
  let i be Integer;
  hereby
    consider k such that
A1: i = k or i = -k by INT_1:2;
    consider m being Element of NAT such that
A2: k = 2*m or k = 2*m+1 by SCHEME1:1;
    assume
A3: i is odd;
    assume
A4: for j being Integer holds i <> 2*j+1;
    per cases by A1,A2;
    suppose
      i = k & k = 2*m;
      hence contradiction by A3,Lm1;
    end;
    suppose
      i = -k & k = 2*m;
      then i = 2*(-m);
      hence contradiction by A3,Lm1;
    end;
    suppose
      i = k & k = 2*m+1;
      hence contradiction by A4;
    end;
    suppose
      i = -k & k = 2*m+1;
      then i = 2*-(m+1)+1;
      hence contradiction by A4;
    end;
  end;
  given j being Integer such that
A5: i = 2*j+1;
  given k being Integer such that
A6: i = 2*k;
  1 = 2*(k - j) by A5,A6;
  hence contradiction by INT_1:9;
end;

registration
  let i be Integer;
  cluster 2*i -> even;
  coherence by Lm1;
end;

registration
  let i be even Integer;
  cluster i+1 -> odd;
  coherence
  proof
    ex j being Integer st i = 2*j by Lm1;
    hence thesis by Th1;
  end;
end;

registration
  let i be odd Integer;
  cluster i+1 -> even;
  coherence
  proof
    consider j being Integer such that
A1: i = 2*j+1 by Th1;
    i+1 = 2*(j+1) by A1;
    hence thesis;
  end;
end;

registration
  let i be even Integer;
  cluster i-1 -> odd;
  coherence
  proof
    consider j being Integer such that
A1: i = 2*j by Lm1;
    i-1 = 2*(j-1)+1 by A1;
    hence thesis;
  end;
end;

registration
  let i be odd Integer;
  cluster i-1 -> even;
  coherence
  proof
    ex j being Integer st i = 2*j+1 by Th1;
    hence thesis;
  end;
end;

registration
  let i be even Integer, j be Integer;
  cluster i*j -> even;
  coherence
  proof
    consider k being Integer such that
A1: i = 2*k by Lm1;
    i*j = 2*(k*j) by A1;
    hence thesis;
  end;
  cluster j*i -> even;
  coherence;
end;

registration
  let i, j be odd Integer;
  cluster i*j -> odd;
  coherence
  proof
    consider l being Integer such that
A1: j = 2*l+1 by Th1;
    consider k being Integer such that
A2: i = 2*k+1 by Th1;
    i*j = 2*(k*(2*l)+(k*1)+(l*1))+1 by A2,A1;
    hence thesis;
  end;
end;

registration
  let i, j be even Integer;
  cluster i+j -> even;
  coherence
  proof
    consider l being Integer such that
A1: j = 2*l by Lm1;
    consider k being Integer such that
A2: i = 2*k by Lm1;
    i+j = 2*(k+l) by A2,A1;
    hence thesis;
  end;
end;

registration
  let i be even Integer, j be odd Integer;
  cluster i+j -> odd;
  coherence
  proof
    consider l being Integer such that
A1: j = 2*l+1 by Th1;
    consider k being Integer such that
A2: i = 2*k by Lm1;
    i+j = 2*(k+l)+1 by A2,A1;
    hence thesis;
  end;
  cluster j+i -> odd;
  coherence;
end;

registration
  let i, j be odd Integer;
  cluster i+j -> even;
  coherence
  proof
    consider l being Integer such that
A1: j = 2*l+1 by Th1;
    consider k being Integer such that
A2: i = 2*k+1 by Th1;
    j+i = 2*(k+l+1) by A2,A1;
    hence thesis;
  end;
end;

registration
  let i be even Integer, j be odd Integer;
  cluster i-j -> odd;
  coherence
  proof
    consider l being Integer such that
A1: j = 2*l+1 by Th1;
    consider k being Integer such that
A2: i = 2*k by Lm1;
    i-j = 2*(k-l)-1 by A2,A1;
    hence thesis;
  end;
  cluster j-i -> odd;
  coherence
  proof
    consider l being Integer such that
A3: j = 2*l+1 by Th1;
    consider k being Integer such that
A4: i = 2*k by Lm1;
    j-i = 2*(l-k)+1 by A4,A3;
    hence thesis;
  end;
end;

registration
  let i, j be odd Integer;
  cluster i-j -> even;
  coherence
  proof
    consider l being Integer such that
A1: j = 2*l+1 by Th1;
    consider k being Integer such that
A2: i = 2*k+1 by Th1;
    i-j = 2*(k-l) by A2,A1;
    hence thesis;
  end;
end;

registration
  let m be even Integer;
  cluster m + 2 -> even;
  coherence
  proof
    2 = 2*1;
    then reconsider t = 2 as even Integer;
    m + t is even;
    hence thesis;
  end;
end;

registration
  let m be odd Integer;
  cluster m + 2 -> odd;
  coherence
  proof
    2 = 2*1;
    then reconsider t = 2 as even Integer;
    m + t is odd;
    hence thesis;
  end;
end;

definition
  let E, f; let n be Nat;
  redefine func iter(f, n) -> Function of E, E;
  coherence by FUNCT_7:83;
end;

theorem Th2:
  for S being non empty Subset of NAT st 0 in S holds min S = 0
by XXREAL_2:def 7;

theorem Th3:
  for E being non empty set, f being Function of E, E, x being
  Element of E holds iter(f,0).x = x
proof
  let E be non empty set, f be Function of E, E, x be Element of E;
  dom f = E by FUNCT_2:def 1;
  then
A1: x in dom f \/ rng f by XBOOLE_0:def 3;
  thus iter(f,0).x = id(field f).x by FUNCT_7:68
    .= x by A1,FUNCT_1:17;
end;

:: from KNASTER, 2005.02.06, A.T.

definition
  let x be object, f be Function;
  pred x is_a_fixpoint_of f means

  x in dom f & x = f.x;
end;

definition
  let A be non empty set, a be Element of A, f be Function of A, A;
  redefine pred a is_a_fixpoint_of f means

  a = f.a;
  compatibility
  proof
    thus a is_a_fixpoint_of f implies a = f.a;
    assume
A1: a = f.a;
    a in A;
    hence a in dom f by FUNCT_2:52;
    thus a = f.a by A1;
  end;
end;

definition
  let f be Function;
  attr f is with_fixpoint means

  ex x being object st x is_a_fixpoint_of f;
end;

notation
  let f be Function;
  antonym f is without_fixpoints for f is with_fixpoint;
end;

definition
  let X be set, x be Element of X;
  attr x is covering means

  union x = union union X;
end;

theorem Th4:
  sE is covering iff union sE = E
proof
  union union bool bool E = union bool E by ZFMISC_1:81
    .= E by ZFMISC_1:81;
  hence thesis;
end;

registration
  let E;
  cluster non empty finite covering for Subset-Family of E;
  existence
  proof
    reconsider sE = {E} as Subset-Family of E by ZFMISC_1:68;
    take sE;
    thus sE is non empty finite;
    union sE = E by ZFMISC_1:25;
    hence thesis by Th4;
  end;
end;

theorem
  for E being set, f being Function of E, E, sE being non empty covering
  Subset-Family of E st for X being Element of sE holds X misses f.:X holds
  f is without_fixpoints
proof
  let E be set, f be Function of E, E, sE be non empty covering Subset-Family
  of E;
  assume
A1: for X being Element of sE holds X misses f.:X;
  given x being object such that
A2: x is_a_fixpoint_of f;
A3: f.x = x by A2;
A4: x in dom f by A2;
  dom f = E by FUNCT_2:52;
  then x in union sE by A4,Th4;
  then consider X being set such that
A5: x in X and
A6: X in sE by TARSKI:def 4;
  f.x in f.:X by A4,A5,FUNCT_1:def 6;
  then X meets f.:X by A3,A5,XBOOLE_0:3;
  hence contradiction by A1,A6;
end;

definition
  let E, f;
  func =_f -> Equivalence_Relation of E means
  :Def7:
  for x, y st x in E & y in
  E holds [x,y] in it iff ex k, l st iter(f,k).x = iter(f,l).y;
  existence
  proof
    defpred P[object,object] means
     $1 in E & $2 in E & ex k, l st iter(f,k).$1 = iter(f,l).$2;
A1: now
      let x be object;
A2:   iter(f,0).x = iter(f,0).x;
      assume x in E;
      hence P[x,x] by A2;
    end;
A3: now
      let x,y,z be object;
      assume that
A4:   P[x,y] and
A5:   P[y,z];
      consider k, l such that
A6:   iter(f,k).x = iter(f,l).y by A4;
      consider m, n such that
A7:   iter(f,m).y =iter(f,n).z by A5;
A8:   dom iter(f,m) = E by FUNCT_2:52;
A9:   dom iter(f,l) = E by FUNCT_2:52;
A10:  dom iter(f,k) = E by FUNCT_2:52;
A11:  dom iter(f,n) = E by FUNCT_2:52;
      iter(f,k+m).x = (iter(f,m)*iter(f,k)).x by FUNCT_7:77
        .= iter(f,m).(iter(f,l).y) by A4,A6,A10,FUNCT_1:13
        .= (iter(f,m)*(iter(f,l))).y by A4,A9,FUNCT_1:13
        .= (iter(f,m+l)).y by FUNCT_7:77
        .= (iter(f,l)*iter(f,m)).y by FUNCT_7:77
        .= iter(f,l).(iter(f,n).z) by A4,A7,A8,FUNCT_1:13
        .= (iter(f,l)*iter(f,n)).z by A5,A11,FUNCT_1:13
        .= iter(f,l+n).z by FUNCT_7:77;
      hence P[x,z] by A4,A5;
    end;
A12: for x,y being object st P[x,y] holds P[y,x];
    consider EqR being Equivalence_Relation of E such that
A13: for x,y being object holds [x,y] in EqR iff x in E & y in E & P[x,y]
from
    EQREL_1:sch 1(A1, A12, A3);
    take EqR;
    let x, y;
    assume x in E & y in E;
    hence thesis by A13;
  end;
  uniqueness
  proof
    let IT1, IT2 be Equivalence_Relation of E such that
A14: for x, y st x in E & y in E holds [x,y] in IT1 iff ex k, l st
    iter(f,k).x = iter(f,l).y and
A15: for x, y st x in E & y in E holds [x,y] in IT2 iff ex k, l st
    iter(f,k).x = iter(f,l).y;
    let a, b be object;
    hereby
      assume
A16:  [a, b] in IT1;
      then
A17:  a in E & b in E by ZFMISC_1:87;
      then ex k, l st iter(f,k).a = iter(f,l).b by A14,A16;
      hence [a, b] in IT2 by A15,A17;
    end;
    assume
A18: [a, b] in IT2;
    then
A19: a in E & b in E by ZFMISC_1:87;
    then ex k, l st iter(f,k).a = iter(f,l).b by A15,A18;
    hence thesis by A14,A19;
  end;
end;

theorem Th6:
  for E being non empty set, f being Function of E, E, c being
  Element of Class =_f, e being Element of c holds f.e in c
proof
  let E be non empty set, f be Function of E, E;
  let c be Element of Class =_f, e be Element of c;
  dom f = E by FUNCT_2:def 1;
  then
A1: f.e in dom f \/ rng f by XBOOLE_0:def 3;
  ex x9 being object st x9 in E & c = Class(=_f, x9) by EQREL_1:def 3;
  then
A2: c = Class(=_f, e) by EQREL_1:23;
  iter(f, 1).e = f.e by FUNCT_7:70
    .= id(field f).(f.e) by A1,FUNCT_1:17
    .= iter(f, 0).(f.e) by FUNCT_7:68;
  then [f.e,e] in =_f by Def7;
  hence thesis by A2,EQREL_1:19;
end;

theorem Th7:
  for E being non empty set, f being Function of E, E, c being
  Element of Class =_f, e being Element of c, n holds iter(f, n).e in c
proof
  let E be non empty set, f be Function of E, E;
  let c be Element of Class =_f, e be Element of c, n;
  dom f = E by FUNCT_2:def 1;
  then iter(f,n).e in dom f \/ rng f by XBOOLE_0:def 3;
  then iter(f, n).e = id(field f).(iter(f,n).e) by FUNCT_1:17
    .= iter(f, 0).(iter(f,n).e) by FUNCT_7:68;
  then
A1: [iter(f,n).e,e] in =_f by Def7;
  ex x9 being object st x9 in E & c = Class(=_f, x9) by EQREL_1:def 3;
  then c = Class(=_f, e) by EQREL_1:23;
  hence thesis by A1,EQREL_1:19;
end;

registration
  cluster empty-membered -> trivial for set;
  coherence;
end;

registration
  let A be set, B be with_non-empty_element set;
  cluster non-empty for Function of A, B;
  existence
  proof
    consider X being non empty set such that
A1: X in B by SETFAM_1:def 10;
    reconsider f = A --> X as Function of A, B by A1,FUNCOP_1:45;
    take f;
    let n be object;
    assume n in dom f;
    then n in A by FUNCOP_1:13;
    hence thesis by FUNCOP_1:7;
  end;
end;

registration
  let A be non empty set, B be with_non-empty_element set, f be non-empty
  Function of A, B, a be Element of A;
  cluster f.a -> non empty;
  coherence
  proof
    dom f = A by FUNCT_2:def 1;
    then f.a in rng f by FUNCT_1:def 3;
    hence thesis;
  end;
end;

registration
  let X be non empty set;
  cluster bool X -> with_non-empty_element;
  coherence
  proof
    take X;
    thus thesis by ZFMISC_1:def 1;
  end;
end;

theorem
  for E being non empty set, f being Function of E, E st f
  is without_fixpoints ex E1, E2, E3 st E1 \/ E2 \/ E3 = E &
  f.:E1 misses E1 & f.:E2 misses E2 & f.:E3 misses E3
proof
  let E be non empty set, f be Function of E, E;
  defpred P[set,Element of [:bool E qua set, bool E, bool E:]] means
$2`1_3 \/ $2`2_3 \/ $2`3_3 = $1 &
  f.:($2`1_3) misses $2`1_3 & f.:($2`2_3) misses $2`2_3 & f.:($2`3_3)
  misses $2`3_3;
  deffunc i(Nat) = iter(f, $1);
  assume
A1: f is without_fixpoints;
A2: for a being Element of Class =_f ex b being Element of [:bool E, bool E,
  bool E:] st P[a,b]
  proof
    reconsider c3 = {} as Subset of E by XBOOLE_1:2;
    let c be Element of Class =_f;
    consider x0 being object such that
A3: x0 in E and
A4: c = Class(=_f, x0) by EQREL_1:def 3;
    reconsider x0 as Element of c by A3,A4,EQREL_1:20;
    defpred P[set] means ex k, l st i(k).$1 = i(l).x0 & k is even & l is even;
    set c1 = { x where x is Element of c : P[x] };
    c1 is Subset of c from DOMAIN_1:sch 7;
    then reconsider c1 as Subset of E by XBOOLE_1:1;
    defpred P[set] means ex k, l st i(k).$1 = i(l).x0 & k is odd & l is even;
    set c2 = { x where x is Element of c : P[x] };
    c2 is Subset of c from DOMAIN_1:sch 7;
    then reconsider c2 as Subset of E by XBOOLE_1:1;
    per cases;
    suppose
A5:   c1 misses c2;
      take b = [c1,c2,c3];
A6:   b`2_3 = c2 by MCART_1:def 6;
A7:   b`3_3 = c3 by MCART_1:def 7;
A8:   b`1_3 = c1 by MCART_1:def 5;
      thus b`1_3 \/ b`2_3 \/ b`3_3 = c
      proof
        hereby
          let x be object;
          assume
A9:       x in b`1_3 \/ b`2_3 \/ b`3_3;
          per cases by A8,A6,A7,A9,XBOOLE_0:def 3;
          suppose
            x in c1;
            then
            ex xx being Element of c st x = xx & ex k, l st i(k).xx = i(l)
            .x0 & k is even & l is even;
            hence x in c;
          end;
          suppose
            x in c2;
            then
            ex xx being Element of c st x = xx & ex k, l st i(k).xx = i(l)
            .x0 & k is odd & l is even;
            hence x in c;
          end;
          suppose
            x in c3;
            hence x in c;
          end;
        end;
        let x be object;
        assume x in c;
        then reconsider xc = x as Element of c;
        [xc,x0] in =_f by A4,EQREL_1:19;
        then consider k, l such that
A10:    i(k).xc = i(l).x0 by Def7;
A11:    dom i(l) = E by FUNCT_2:def 1;
A12:    dom i(k) = E by FUNCT_2:def 1;
        per cases;
        suppose
A13:      k is even;
          then reconsider k as even Nat;
          thus x in b`1_3 \/ b`2_3 \/ b`3_3
          proof
            per cases;
            suppose
              l is even;
              then xc in c1 by A10,A13;
              hence thesis by A8,A7,XBOOLE_0:def 3;
            end;
            suppose
              l is odd;
              then reconsider l as odd Nat;
              i(k+1).xc = (f*i(k)).xc by FUNCT_7:71
                .= f.(i(l).x0) by A10,A12,FUNCT_1:13
                .= (f*i(l)).x0 by A11,FUNCT_1:13
                .= i(l+1).x0 by FUNCT_7:71;
              then xc in c2;
              hence thesis by A6,A7,XBOOLE_0:def 3;
            end;
          end;
        end;
        suppose
A14:      k is odd;
          then reconsider k as odd Nat;
          thus x in b`1_3 \/ b`2_3 \/ b`3_3
          proof
            per cases;
            suppose
              l is even;
              then xc in c2 by A10,A14;
              hence thesis by A6,A7,XBOOLE_0:def 3;
            end;
            suppose
              l is odd;
              then reconsider l as odd Nat;
              i(k+1).xc = (f*i(k)).xc by FUNCT_7:71
                .= f.(i(l).x0) by A10,A12,FUNCT_1:13
                .= (f*i(l)).x0 by A11,FUNCT_1:13
                .= i(l+1).x0 by FUNCT_7:71;
              then xc in c1;
              hence thesis by A8,A7,XBOOLE_0:def 3;
            end;
          end;
        end;
      end;
      f.:c1 c= c2
      proof
        let y be object;
A15:    dom f = E by FUNCT_2:def 1;
        assume y in f.:c1;
        then consider x being object such that
        x in dom f and
A16:    x in c1 and
A17:    y = f.x by FUNCT_1:def 6;
        consider xx being Element of c such that
A18:    x = xx and
A19:    ex k, l st i(k).xx = i(l).x0 & k is even & l is even by A16;
        consider k, l such that
A20:    i(k).xx = i(l).x0 and
A21:    k is even & l is even by A19;
        reconsider k, l as even Nat by A21;
        reconsider k1 = k+1 as odd Element of NAT;
        reconsider l1 = l+1 as odd Element of NAT;
        reconsider l2 = l1+1 as even Element of NAT;
A22:    dom i(k) = E by FUNCT_2:def 1;
        reconsider yc = y as Element of c by A17,A18,Th6;
A23:    dom i(l) = E by FUNCT_2:def 1;
A24:    dom i(k1) = E by FUNCT_2:def 1;
A25:    i(k1+1).xx = (i(k1)*f).xx by FUNCT_7:69
          .= i(k1).(f.xx) by A15,FUNCT_1:13;
A26:    dom i(l1) = E by FUNCT_2:def 1;
        i(k1+1).xx = (f*i(k1)).xx by FUNCT_7:71
          .= f.(i(k1).xx) by A24,FUNCT_1:13
          .= f.((f*i(k)).xx) by FUNCT_7:71
          .= f.(f.(i(l).x0)) by A20,A22,FUNCT_1:13
          .= f.((f*i(l)).x0) by A23,FUNCT_1:13
          .= f.(i(l1).x0) by FUNCT_7:71
          .= (f*i(l1)).x0 by A26,FUNCT_1:13
          .= i(l2).x0 by FUNCT_7:71;
        then yc in c2 by A17,A18,A25;
        hence thesis;
      end;
      hence f.:(b`1_3) misses b`1_3 by A5,A8,XBOOLE_1:63;
      f.:c2 c= c1
      proof
        let y be object;
A27:    dom f = E by FUNCT_2:def 1;
        assume y in f.:c2;
        then consider x being object such that
        x in dom f and
A28:    x in c2 and
A29:    y = f.x by FUNCT_1:def 6;
        consider xx being Element of c such that
A30:    x = xx and
A31:    ex k, l st i(k).xx = i(l).x0 & k is odd & l is even by A28;
        consider k, l such that
A32:    i(k).xx = i(l).x0 and
A33:    k is odd and
A34:    l is even by A31;
        reconsider l as even Nat by A34;
        reconsider k as odd Nat by A33;
        reconsider k1 = k+1 as even Element of NAT;
        reconsider l1 = l+1 as odd Element of NAT;
        reconsider l2 = l1+1 as even Element of NAT;
A35:    dom i(k) = E by FUNCT_2:def 1;
        reconsider yc = y as Element of c by A29,A30,Th6;
A36:    dom i(l) = E by FUNCT_2:def 1;
A37:    dom i(k1) = E by FUNCT_2:def 1;
A38:    i(k1+1).xx = (i(k1)*f).xx by FUNCT_7:69
          .= i(k1).(f.xx) by A27,FUNCT_1:13;
A39:    dom i(l1) = E by FUNCT_2:def 1;
        i(k1+1).xx = (f*i(k1)).xx by FUNCT_7:71
          .= f.(i(k1).xx) by A37,FUNCT_1:13
          .= f.((f*i(k)).xx) by FUNCT_7:71
          .= f.(f.(i(l).x0)) by A32,A35,FUNCT_1:13
          .= f.((f*i(l)).x0) by A36,FUNCT_1:13
          .= f.(i(l1).x0) by FUNCT_7:71
          .= (f*i(l1)).x0 by A39,FUNCT_1:13
          .= i(l2).x0 by FUNCT_7:71;
        then yc in c1 by A29,A30,A38;
        hence thesis;
      end;
      hence f.:(b`2_3) misses b`2_3 by A5,A6,XBOOLE_1:63;
      thus thesis by A7;
    end;
    suppose
      c1 meets c2;
      then consider x1 being object such that
A40:  x1 in c1 and
A41:  x1 in c2 by XBOOLE_0:3;
      consider x11 being Element of c such that
A42:  x1 = x11 and
A43:  ex k, l st i(k).x11 = i(l).x0 & k is even & l is even by A40;
      consider x12 being Element of c such that
A44:  x1 = x12 and
A45:  ex k, l st i(k).x12 = i(l).x0 & k is odd & l is even by A41;
      consider k2, l2 being Nat such that
A46:  i(k2).x12 = i(l2).x0 and
A47:  k2 is odd and
A48:  l2 is even by A45;
      reconsider x1 as Element of c by A42;
      consider k1, l1 being Nat such that
A49:  i(k1).x11 = i(l1).x0 and
A50:  k1 is even & l1 is even by A43;
A51:  dom i(k1) = E by FUNCT_2:def 1;
A52:  dom i(l1) = E by FUNCT_2:def 1;
A53:  i(l2+k1).x1 = (i(l2)*i(k1)).x1 by FUNCT_7:77
        .= i(l2).(i(l1).x0) by A42,A49,A51,FUNCT_1:13
        .= (i(l2)*i(l1)).x0 by A52,FUNCT_1:13
        .= i(l1+l2).x0 by FUNCT_7:77;
A54:  dom i(l2) = E by FUNCT_2:def 1;
A55:  dom i(k2) = E by FUNCT_2:def 1;
A56:  i(l1+k2).x1 = (i(l1)*i(k2)).x1 by FUNCT_7:77
        .= i(l1).(i(l2).x0) by A44,A46,A55,FUNCT_1:13
        .= (i(l1)*i(l2)).x0 by A54,FUNCT_1:13
        .= i(l1+l2).x0 by FUNCT_7:77;
      ex r being Element of E, k being odd Element of NAT st i(k).r = r &
      r in c
      proof
        reconsider k2 as odd Nat by A47;
        reconsider k1, l1, l2 as even Nat by A50,A48;
A57:    dom i(k1+l2) = E by FUNCT_2:def 1;
A58:    dom i(k2+l1) = E by FUNCT_2:def 1;
        per cases by XXREAL_0:1;
        suppose
          k1+l2 < k2+l1;
          then reconsider k = k2+l1-(k1+l2) as Element of NAT by INT_1:5;
          take r = i(k1+l2).x1;
          reconsider k as odd Element of NAT;
          take k;
          i(k).(i(k1+l2).x1) = (i(k)*i(k1+l2)).x1 by A57,FUNCT_1:13
            .= i(k+(k1+l2)).x1 by FUNCT_7:77
            .= i(k1+l2).x1 by A56,A53;
          hence i(k).r = r;
          thus thesis by Th7;
        end;
        suppose
          k1+l2 > k2+l1;
          then reconsider k = k1+l2-(k2+l1) as Element of NAT by INT_1:5;
          take r = i(k2+l1).x1;
          reconsider k as odd Element of NAT;
          take k;
          i(k).(i(k2+l1).x1) = (i(k)*i(k2+l1)).x1 by A58,FUNCT_1:13
            .= i(k+(k2+l1)).x1 by FUNCT_7:77
            .= i(k2+l1).x1 by A56,A53;
          hence i(k).r = r;
          thus thesis by Th7;
        end;
      end;
      then consider r being Element of E, k being odd Element of NAT such that
A59:  i(k).r = r and
A60:  r in c;
      reconsider r as Element of c by A60;
      deffunc F(set) = {l where l is Element of NAT : i(l).$1 = r};
A61:  for x being Element of c holds F(x) in bool NAT
      proof
        let x be Element of c;
        defpred P1[Element of NAT] means i($1).x = r;
        { l where l is Element of NAT : P1[l]} is Subset of NAT from
        DOMAIN_1:sch 7;
        hence thesis;
      end;
      consider Odl being Function of c, bool NAT such that
A62:  for x being Element of c holds Odl.x = F(x) from FUNCT_2:sch 8(
      A61);
      now
        defpred P[Nat] means i(k*$1).r = r;
        let n be object;
        assume n in dom Odl;
        then reconsider nc = n as Element of c by FUNCT_2:def 1;
A63:    Odl.nc = {l where l is Element of NAT : i(l).nc = r} by A62;
A64:    now
          let i be Nat;
          assume
A65:      P[i];
A66:      dom i(k) = E by FUNCT_2:def 1;
          i(k*(i+1)).r = i(k*i+k*1).r .= (i(k*i)*i(k)).r by FUNCT_7:77
            .= r by A59,A65,A66,FUNCT_1:13;
          hence P[i+1];
        end;
A67:    P[0] by Th3;
A68:    for i being Nat holds P[i] from NAT_1:sch 2(A67,A64);
        ex x9 being object st x9 in E & c = Class(=_f, x9) by EQREL_1:def 3;
        then [nc, r] in =_f by EQREL_1:22;
        then consider kn, ln being Nat such that
A69:    iter(f,kn).nc = iter(f,ln).r by Def7;
A70:    dom i(ln) = E by FUNCT_2:def 1;
        set mk = ln mod k;
        set dk = ln div k;
A71:    dom i(kn) = E by FUNCT_2:def 1;
A72:    2*0 <> k;
        then mk < k by NAT_D:1;
        then reconsider kmk = k - mk as Element of NAT by INT_1:5;
        ln = k*dk+mk by A72,NAT_D:2;
        then
A73:    ln+kmk = k*(dk+1);
        i(kmk+kn).nc = (i(kmk)*i(kn)).nc by FUNCT_7:77
          .= i(kmk).(i(ln).r) by A69,A71,FUNCT_1:13
          .= (i(kmk)*i(ln)).r by A70,FUNCT_1:13
          .= i(k*(dk+1)).r by A73,FUNCT_7:77
          .= r by A68;
        then kn+kmk in Odl.n by A63;
        hence Odl.n is non empty;
      end;
      then reconsider Odl as non-empty Function of c, bool NAT by FUNCT_1:def 9
;
      deffunc F(Element of c) = min (Odl.$1);
      consider odl being Function of c, NAT such that
A74:  for x being Element of c holds odl.x = F(x) from FUNCT_2:sch 4;
      defpred P1[Element of c] means odl.$1 is even;
      set c1 = { x where x is Element of c : P1[x]};
      set d1 = c1 \ {r};
      c1 is Subset of c from DOMAIN_1:sch 7;
      then
A75:  d1 c= c by XBOOLE_1:1;
      i(0).r = r by Th3;
      then 0 in {l where l is Element of NAT : i(l).r = r};
      then 0 in Odl.r by A62;
      then min (Odl.r) = 0 by Th2;
      then
A76:  odl.r = 2*0 by A74;
      then
A77:  r in c1;
      reconsider d1 as Subset of E by A75,XBOOLE_1:1;
      defpred P2[Element of c] means odl.$1 is odd;
      set d2 = { x where x is Element of c : P2[x]};
      d2 is Subset of c from DOMAIN_1:sch 7;
      then reconsider d2 as Subset of E by XBOOLE_1:1;
A78:  for x being Element of c st x <> r holds odl.(f.x) = (odl.x qua
      Element of NAT)-1
      proof
        let x be Element of c;
        reconsider fx = f.x as Element of c by Th6;
        reconsider ofx = odl.(fx), ox = odl.x as Element of NAT;
        assume
A79:    x <> r;
        now
          assume odl.x = 0;
          then 0 = min (Odl.x) by A74;
          then 0 in Odl.x by XXREAL_2:def 7;
          then 0 in {l where l is Element of NAT : i(l).x = r} by A62;
          then ex l being Element of NAT st l = 0 & i(l).x = r;
          hence contradiction by A79,Th3;
        end;
        then reconsider ox1 = ox-1 as Element of NAT by INT_1:5,NAT_1:14;
        ox = min (Odl.x) by A74;
        then ox in Odl.x by XXREAL_2:def 7;
        then ox in {l where l is Element of NAT : i(l).x = r} by A62;
        then
A80:    ex l being Element of NAT st l = ox & i(l).x = r;
A81:    dom f = E by FUNCT_2:def 1;
        then i(ox1).fx = (i(ox1)*f).x by FUNCT_1:13
          .= i(ox1+1).x by FUNCT_7:69
          .= i(ox).x;
        then ox1 in {l where l is Element of NAT : i(l).fx = r} by A80;
        then
A82:    ox1 in Odl.fx by A62;
        ofx = min (Odl.fx) by A74;
        then ofx in Odl.fx by XXREAL_2:def 7;
        then ofx in {l where l is Element of NAT : i(l).fx = r} by A62;
        then
A83:    ex l being Element of NAT st l = ofx & i(l).fx = r;
        i(ofx+1).x = (i(ofx)*f).x by FUNCT_7:69
          .= i(ofx).fx by A81,FUNCT_1:13;
        then ofx+1 in {l where l is Element of NAT : i(l).x = r} by A83;
        then
A84:    ofx+1 in Odl.x by A62;
        ox = min (Odl.x) by A74;
        then ofx+1 >= ox by A84,XXREAL_2:def 7;
        then
A85:    ofx >= ox-1 by XREAL_1:20;
        ofx = min (Odl.fx) by A74;
        then ofx <= ox-1 by A82,XXREAL_2:def 7;
        hence thesis by A85,XXREAL_0:1;
      end;
A86:  f.:d1 c= d2
      proof
        let y be object;
        assume y in f.:d1;
        then consider x being object such that
        x in dom f and
A87:    x in d1 and
A88:    y = f.x by FUNCT_1:def 6;
        x in c1 by A87;
        then consider xx being Element of c such that
A89:    x = xx and
A90:    odl.xx is even;
        reconsider ox = odl.xx as even Element of NAT by A90;
        reconsider yc = y as Element of c by A88,A89,Th6;
        r <> xx by A87,A89,ZFMISC_1:56;
        then odl.yc = ox-1 by A78,A88,A89;
        hence thesis;
      end;
A91:  c1 \/ d2 = c
      proof
        hereby
          let x be object;
          assume
A92:      x in c1 \/ d2;
          per cases by A92,XBOOLE_0:def 3;
          suppose
            x in c1;
            then ex xc being Element of c st xc = x & odl.xc is even;
            hence x in c;
          end;
          suppose
            x in d2;
            then ex xc being Element of c st xc = x & odl.xc is odd;
            hence x in c;
          end;
        end;
        let x be object;
        assume x in c;
        then reconsider xc = x as Element of c;
        odl.xc is even or odl.xc is odd;
        then x in c1 or x in d2;
        hence thesis by XBOOLE_0:def 3;
      end;
      reconsider d3 = {r} as Subset of E by ZFMISC_1:31;
      take b = [d1,d2,d3];
A93:  b`1_3 = d1 by MCART_1:def 5;
A94:  b`2_3 = d2 by MCART_1:def 6;
A95:  b`3_3 = d3 by MCART_1:def 7;
      d1 \/ d3 = c1 \/ d3 by XBOOLE_1:39
        .= c1 by A77,ZFMISC_1:40;
      hence b`1_3 \/ b`2_3 \/ b`3_3 = c by A93,A94,A95,A91,XBOOLE_1:4;
A96:  c1 misses d2
      proof
        assume c1 meets d2;
        then consider z being object such that
A97:    z in c1 & z in d2 by XBOOLE_0:3;
        ( ex x being Element of c st z = x & odl.x is even)& ex x being
        Element of c st z = x & odl.x is odd by A97;
        hence contradiction;
      end;
      then d1 misses d2 by XBOOLE_1:63;
      hence f.:(b`1_3) misses b`1_3 by A93,A86,XBOOLE_1:63;
      f.:d2 c= c1
      proof
        let y be object;
        assume y in f.:d2;
        then consider x being object such that
        x in dom f and
A98:    x in d2 and
A99:    y = f.x by FUNCT_1:def 6;
        consider xx being Element of c such that
A100:   x = xx and
A101:   odl.xx is odd by A98;
        reconsider ox = odl.xx as odd Element of NAT by A101;
        reconsider yc = y as Element of c by A99,A100,Th6;
        odl.yc = ox-1 by A76,A78,A99,A100;
        hence thesis;
      end;
      hence f.:(b`2_3) misses b`2_3 by A94,A96,XBOOLE_1:63;
      thus f.:(b`3_3) misses b`3_3
      proof
        assume f.:(b`3_3) meets b`3_3;
        then consider y being object such that
A102:   y in f.:(b`3_3) and
A103:   y in b`3_3 by XBOOLE_0:3;
A104:   y = r by A95,A103,TARSKI:def 1;
        consider x being object such that
        x in dom f and
A105:   x in {r} and
A106:   y = f.x by A95,A102,FUNCT_1:def 6;
        x = r by A105,TARSKI:def 1;
        then r is_a_fixpoint_of f by A104,A106;
        hence contradiction by A1;
      end;
    end;
  end;
  consider F being Function of Class =_f, [:bool E, bool E, bool E:] such that
A107: for a being Element of Class =_f holds P[a,F.a] from FUNCT_2:sch 3
  (A2);
  set E3c = the set of all (F.c)`3_3 where c is Element of Class =_f;
  set E2c = the set of all (F.c)`2_3 where c is Element of Class =_f;
  set E1c = the set of all (F.c)`1_3 where c is Element of Class =_f;
  set E1 = union E1c;
  set E2 = union E2c;
  set E3 = union E3c;
  take E1, E2, E3;
  thus E1 \/ E2 \/ E3 = E
  proof
    hereby
      let x be object;
      assume x in E1 \/ E2 \/ E3;
      then
A108: x in E1 \/ E2 or x in E3 by XBOOLE_0:def 3;
      per cases by A108,XBOOLE_0:def 3;
      suppose
        x in E1;
        then consider Y being set such that
A109:   x in Y and
A110:   Y in E1c by TARSKI:def 4;
        ex c being Element of Class =_f st Y = (F.c)`1_3 by A110;
        hence x in E by A109;
      end;
      suppose
        x in E2;
        then consider Y being set such that
A111:   x in Y and
A112:   Y in E2c by TARSKI:def 4;
        ex c being Element of Class =_f st Y = (F.c)`2_3 by A112;
        hence x in E by A111;
      end;
      suppose
        x in E3;
        then consider Y being set such that
A113:   x in Y and
A114:   Y in E3c by TARSKI:def 4;
        ex c being Element of Class =_f st Y = (F.c)`3_3 by A114;
        hence x in E by A113;
      end;
    end;
    let x be object;
    set c = Class(=_f,x);
    assume
A115: x in E;
    then
A116: x in c by EQREL_1:20;
    reconsider c as Element of Class =_f by A115,EQREL_1:def 3;
    x in (F.c)`1_3 \/ (F.c)`2_3 \/ (F.c)`3_3 by A107,A116;
    then
A117: x in (F.c)`1_3 \/ (F.c)`2_3 or x in (F.c)`3_3 by XBOOLE_0:def 3;
    per cases by A117,XBOOLE_0:def 3;
    suppose
A118: x in (F.c)`1_3;
      (F.c)`1_3 in E1c;
      then x in E1 by A118,TARSKI:def 4;
      then x in E1 \/ E2 by XBOOLE_0:def 3;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose
A119: x in (F.c)`2_3;
      (F.c)`2_3 in E2c;
      then x in E2 by A119,TARSKI:def 4;
      then x in E1 \/ E2 by XBOOLE_0:def 3;
      hence thesis by XBOOLE_0:def 3;
    end;
    suppose
A120: x in (F.c)`3_3;
      (F.c)`3_3 in E3c;
      then x in E3 by A120,TARSKI:def 4;
      hence thesis by XBOOLE_0:def 3;
    end;
  end;
  thus f.:E1 misses E1
  proof
    assume not thesis;
    then consider x being object such that
A121: x in f.:E1 and
A122: x in E1 by XBOOLE_0:3;
    consider Y being set such that
A123: x in Y and
A124: Y in E1c by A122,TARSKI:def 4;
    consider c being Element of Class =_f such that
A125: Y = (F.c)`1_3 by A124;
    x in (F.c)`1_3 \/ (F.c)`2_3 by A123,A125,XBOOLE_0:def 3;
    then x in (F.c)`1_3 \/ (F.c)`2_3 \/ (F.c)`3_3 by XBOOLE_0:def 3;
    then
A126: x in c by A107;
    ex x9 being object st x9 in E & c = Class(=_f, x9) by EQREL_1:def 3;
    then
A127: c = Class(=_f, x) by A126,EQREL_1:23;
    dom f = E by FUNCT_2:def 1;
    then
A128: x in dom f \/ rng f by A123,A125,XBOOLE_0:def 3;
    consider xx being object such that
A129: xx in dom f and
A130: xx in E1 and
A131: x = f.xx by A121,FUNCT_1:def 6;
    consider YY being set such that
A132: xx in YY and
A133: YY in E1c by A130,TARSKI:def 4;
    consider cc being Element of Class =_f such that
A134: YY = (F.cc)`1_3 by A133;
    xx in (F.cc)`1_3 \/ (F.cc)`2_3 by A132,A134,XBOOLE_0:def 3;
    then xx in (F.cc)`1_3 \/ (F.cc)`2_3 \/ (F.cc)`3_3 by XBOOLE_0:def 3;
    then
A135: xx in cc by A107;
    ex xx9 being object st xx9 in E & cc = Class(=_f, xx9) by EQREL_1:def 3;
    then
A136: cc = Class(=_f, xx) by A135,EQREL_1:23;
    iter(f, 1).xx = x by A131,FUNCT_7:70
      .= id(field f).x by A128,FUNCT_1:17
      .= iter(f, 0).x by FUNCT_7:68;
    then [x,xx] in =_f by A123,A125,A132,A134,Def7;
    then
A137: Class(=_f, x) = Class(=_f, xx) by A123,A125,EQREL_1:35;
A138: f.xx in f.:YY by A129,A132,FUNCT_1:def 6;
    f.:YY misses YY by A107,A134;
    hence contradiction by A123,A125,A131,A134,A127,A136,A137,A138,XBOOLE_0:3;
  end;
  thus f.:E2 misses E2
  proof
    assume not thesis;
    then consider x being object such that
A139: x in f.:E2 and
A140: x in E2 by XBOOLE_0:3;
    consider Y being set such that
A141: x in Y and
A142: Y in E2c by A140,TARSKI:def 4;
    consider c being Element of Class =_f such that
A143: Y = (F.c)`2_3 by A142;
    x in (F.c)`1_3 \/ (F.c)`2_3 by A141,A143,XBOOLE_0:def 3;
    then x in (F.c)`1_3 \/ (F.c)`2_3 \/ (F.c)`3_3 by XBOOLE_0:def 3;
    then
A144: x in c by A107;
    ex x9 being object st x9 in E & c = Class(=_f, x9) by EQREL_1:def 3;
    then
A145: c = Class(=_f, x) by A144,EQREL_1:23;
    dom f = E by FUNCT_2:def 1;
    then
A146: x in dom f \/ rng f by A141,A143,XBOOLE_0:def 3;
    consider xx being object such that
A147: xx in dom f and
A148: xx in E2 and
A149: x = f.xx by A139,FUNCT_1:def 6;
    consider YY being set such that
A150: xx in YY and
A151: YY in E2c by A148,TARSKI:def 4;
    consider cc being Element of Class =_f such that
A152: YY = (F.cc)`2_3 by A151;
    xx in (F.cc)`1_3 \/ (F.cc)`2_3 by A150,A152,XBOOLE_0:def 3;
    then xx in (F.cc)`1_3 \/ (F.cc)`2_3 \/ (F.cc)`3_3 by XBOOLE_0:def 3;
    then
A153: xx in cc by A107;
    ex xx9 being object st xx9 in E & cc = Class(=_f, xx9) by EQREL_1:def 3;
    then
A154: cc = Class(=_f, xx) by A153,EQREL_1:23;
    iter(f, 1).xx = x by A149,FUNCT_7:70
      .= id(field f).x by A146,FUNCT_1:17
      .= iter(f, 0).x by FUNCT_7:68;
    then [x,xx] in =_f by A141,A143,A150,A152,Def7;
    then
A155: Class(=_f, x) = Class(=_f, xx) by A141,A143,EQREL_1:35;
A156: f.xx in f.:YY by A147,A150,FUNCT_1:def 6;
    f.:YY misses YY by A107,A152;
    hence contradiction by A141,A143,A149,A152,A145,A154,A155,A156,XBOOLE_0:3;
  end;
  thus f.:E3 misses E3
  proof
    assume not thesis;
    then consider x being object such that
A157: x in f.:E3 and
A158: x in E3 by XBOOLE_0:3;
    consider Y being set such that
A159: x in Y and
A160: Y in E3c by A158,TARSKI:def 4;
    consider c being Element of Class =_f such that
A161: Y = (F.c)`3_3 by A160;
    x in (F.c)`1_3 \/ (F.c)`2_3 \/ (F.c)`3_3 by A159,A161,XBOOLE_0:def 3;
    then
A162: x in c by A107;
    ex x9 being object st x9 in E & c = Class(=_f, x9) by EQREL_1:def 3;
    then
A163: c = Class(=_f, x) by A162,EQREL_1:23;
    dom f = E by FUNCT_2:def 1;
    then
A164: x in dom f \/ rng f by A159,A161,XBOOLE_0:def 3;
    consider xx being object such that
A165: xx in dom f and
A166: xx in E3 and
A167: x = f.xx by A157,FUNCT_1:def 6;
    consider YY being set such that
A168: xx in YY and
A169: YY in E3c by A166,TARSKI:def 4;
    consider cc being Element of Class =_f such that
A170: YY = (F.cc)`3_3 by A169;
    xx in (F.cc)`1_3 \/ (F.cc)`2_3 \/ (F.cc)`3_3 by A168,A170,XBOOLE_0:def 3;
    then
A171: xx in cc by A107;
    ex xx9 being object st xx9 in E & cc = Class(=_f, xx9) by EQREL_1:def 3;
    then
A172: cc = Class(=_f, xx) by A171,EQREL_1:23;
    iter(f, 1).xx = x by A167,FUNCT_7:70
      .= id(field f).x by A164,FUNCT_1:17
      .= iter(f, 0).x by FUNCT_7:68;
    then [x,xx] in =_f by A159,A161,A168,A170,Def7;
    then
A173: Class(=_f, x) = Class(=_f, xx) by A159,A161,EQREL_1:35;
A174: f.xx in f.:YY by A165,A168,FUNCT_1:def 6;
    f.:YY misses YY by A107,A170;
    hence contradiction by A159,A161,A167,A170,A163,A172,A173,A174,XBOOLE_0:3;
  end;
end;

begin :: Addenda
:: from SCMFSA9A, 2006.03.14, A.T.

theorem
  for n being Nat holds
  n is odd iff ex k being Nat st n = 2*k+1
proof
  let n be Nat;
  hereby
    assume
A1: n is odd;
    then consider j being Integer such that
A2: n = 2*j+1 by Th1;
    now
      assume j < 0;
      then
A3:   2*j + 1 <= 2*0 by INT_1:7,XREAL_1:68;
      per cases by A3;
      suppose
        2*j+1 < 0;
        hence contradiction by A2;
      end;
      suppose
        2*j+1 = 0;
        then n = 2*0;
        hence contradiction by A1;
      end;
    end;
    then j in NAT by INT_1:3;
    then reconsider j as Nat;
    take j;
    thus n = 2*j+1 by A2;
  end;
  thus thesis;
end;

:: missing, 2008.03.20, A.T.

theorem
  for A being non empty set, f being Function of A,A, x being Element of
  A holds iter(f,n+1).x = f.(iter(f,n).x)
proof
  let A be non empty set, f be Function of A,A, x be Element of A;
  thus iter(f,n+1).x = (f*iter(f,n)).x by FUNCT_7:71
    .= f.(iter(f,n).x) by FUNCT_2:15;
end;

theorem
  for i being Integer holds i is even iff ex j being Integer st i = 2*j by Lm1;

:: from HEYTING3, MOEBIUS1, 2010.02.13, A.T.

registration
  cluster odd for Nat;
  existence
  proof
    take 1;
    1 = 2*0+1;
    hence thesis;
  end;
  cluster even for Nat;
  existence
  proof
    take 0;
    0 = 2*0;
    hence thesis;
  end;
end;

theorem Th12:
  for n being odd Nat holds 1 <= n
proof
  let n be odd Nat;
  2 * 0 < n;
  then 0 + 1 <= n by NAT_1:13;
  hence thesis;
end;

registration
 cluster odd -> non zero for Integer;
 coherence by Th12;
end;