hcju/proofwikiqa huggingface dataset:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \paren {a + b} \times 1       | r = a + b       | c = Axiom $\text A$ }} {{eqn | r = \paren {a \times 1} + \paren {b \times 1}       | c = Axiom $\text A$ }} {{end-eqn}} and so $\map P 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$ Then we need to show: :$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {a + b} \times \paren {k + 1}       | r = \paren {\paren {a + b} \times k} + \paren {a + b}       | c = Axiom $\text B$ }} {{eqn | r = \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}       | c = Induction hypothesis }} {{eqn | r = \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}       | c = Natural Number Addition is Commutative }} {{eqn | r = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }       | c = Axiom $\text B$ }} {{end-eqn}} The result follows by the Principle of Mathematical Induction.
Let $A = \set {n \in \N: \map P n}$. We show that $A$ is an inductive set. By $(1)$: :$\forall 1 \le i \le d: i \in A$ Let: :$\forall x \ge d: \set {1, 2, \dotsc, x} \subset A$ Then by definition of $A$: :$\forall k \in \N: x - \paren {d - 1} \le k < x + 1: \map P k$ Thus $\map P {x + 1} \implies x + 1 \in A$ Thus $A$ is an inductive set. Thus by the fifth axiom of Peano: :$\forall n \in \N: A = \N \implies \map P n$ {{qed}} {{Proofread}}
=== Reflexivity === :$\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$ So $\boxtimes$ is a reflexive relation. {{qed|lemma}} === Symmetry === {{begin-eqn}} {{eqn | l = \tuple {x_1, y_1}       | o = \boxtimes       | r = \tuple {x_2, y_2}       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ y_2       | r = x_2 \circ y_1       | c =  }} {{eqn | ll= \leadsto       | l = x_2 \circ y_1       | r = x_1 \circ y_2       | c = $\circ$ is commutative }} {{eqn | ll= \leadsto       | l = \tuple {x_2, y_2}       | o = \boxtimes       | r = \tuple {x_1, y_1}       | c =  }} {{end-eqn}} So $\boxtimes$ is a symmetric relation. {{qed|lemma}} === Transitivity === {{begin-eqn}} {{eqn | l = \tuple {x_1, y_1}       | o = \boxtimes       | r = \tuple {x_2, y_2}       | c =  }} {{eqn | lo= \land       | l = \tuple {x_2, y_2}       | o = \boxtimes       | r = \tuple {x_3, y_3}       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ y_2       | r = x_2 \circ y_1       | c =  }} {{eqn | lo= \land       | l = x_2 \circ y_3       | r = x_3 \circ y_2       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ y_3 \circ y_2       | r = x_1 \circ y_2 \circ y_3       | c = $\circ$ is commutative }} {{eqn | r = x_2 \circ y_1 \circ y_3       | c = substituting $x_2 \circ y_1$ for $x_1 \circ y_2$ }} {{eqn | r = x_2 \circ y_3 \circ y_1       | c = $\circ$ is commutative }} {{eqn | r = x_3 \circ y_2 \circ y_1       | c = substituting $x_3 \circ y_2$ for $x_2 \circ y_3$ }} {{eqn | r = x_3 \circ y_1 \circ y_2       | c = $\circ$ is commutative }} {{eqn | ll= \leadsto       | l = x_1 \circ y_3       | r = x_3 \circ y_1       | c = as $y_2 \in S_2$, therefore $y_2 \in C$, and so is cancellable for $\circ$ }} {{eqn | ll= \leadsto       | l = \tuple {x_1, y_1}       | o = \boxtimes       | r = \tuple {x_3, y_3}       | c =  }} {{end-eqn}} So $\boxtimes$ is a transitive relation. {{qed|lemma}} All the criteria are therefore seen to hold for $\boxtimes$ to be an equivalence relation. {{qed}}
$\boxtimes$ is an instance of a cross-relation. We also have that Natural Number Addition is Commutative. The result therefore follows from Cross-Relation is Equivalence Relation. {{qed}}
We have that $\struct {\N, +}$ is a commutative semigroup from Natural Number Addition is Commutative. We also have the result Cross-Relation on Natural Numbers is Equivalence Relation. The result follows from the definition of equivalence class. {{qed}}
Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$. Then: {{begin-eqn}} {{eqn | l = \eqclass {a_1, b_1} {}       | r = \eqclass {a_2, b_2} {}       | c = {{Defof|Operation Induced by Direct Product}} }} {{eqn | lo= \land       | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c = {{Defof|Operation Induced by Direct Product}} }} {{eqn | ll= \leadstoandfrom       | l = a_1 + b_2       | r = a_2 + b_1       | c = {{Defof|Cross-Relation}} }} {{eqn | lo= \land       | l = c_1 + d_2       | r = c_2 + d_1       | c = {{Defof|Cross-Relation}} }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \tuple {a_1 + c_1} + \tuple {b_2 + d_2}       | r = \tuple {a_1 + b_2} + \tuple {c_1 + d_2}       | c = Commutativity and Associativity of $+$ }} {{eqn | r = \tuple {a_2 + b_1} + \tuple {c_2 + d_1}       | c = from above: $a_1 + b_2 = a_2 + b_1, c_1 + d_2 = c_2 + d_1$ }} {{eqn | r = \tuple {a_2 + c_2} + \tuple {b_1 + d_1}       | c = Commutativity and associativity of $+$ }} {{eqn | ll= \leadsto       | l = \tuple {a_1 + c_1, b_1 + d_1}       | o = \boxtimes       | r = \tuple {a_2 + c_2, b_2 + d_2}       | c = Definition of $\boxtimes$ }} {{eqn | ll= \leadsto       | l = \tuple {\tuple {a_1, b_1} \oplus \tuple {c_1, d_1} }       | o = \boxtimes       | r = \tuple {\tuple {a_2, b_2} \oplus \tuple {c_2, d_2} }       | c = Definition of $\oplus$ }} {{end-eqn}} {{qed}}
Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$. Then: {{begin-eqn}} {{eqn | l = \eqclass {a_1, b_1} {}       | r = \eqclass {a_2, b_2} {}       | c = {{Defof|Operation Induced by Direct Product}} }} {{eqn | lo= \land       | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c = {{Defof|Operation Induced by Direct Product}} }} {{eqn | ll= \leadstoandfrom       | l = a_1 + b_2       | r = a_2 + b_1       | c = {{Defof|Cross-Relation}} }} {{eqn | lo= \land       | l = c_1 + d_2       | r = c_2 + d_1       | c = {{Defof|Cross-Relation}} }} {{end-eqn}} Both $+$ and $\times$ are commutative and associative on $\N$. Thus: {{begin-eqn}} {{eqn | l = \paren {a_1 \cdot c_1 + b_1 \cdot d_1 + a_2 \cdot d_2 + b_2 \cdot c_2}       | o = +       | r = \paren {a_2 \cdot c_1 + b_2 \cdot c_1 + a_2 \cdot d_1 + b_2 \cdot d_1}       | c =  }} {{eqn | lo= =       | l = \paren {a_1 + b_2} \cdot c_1 + \paren {b_1 + a_2} \cdot d_1       | o = +       | r = a_2 \cdot \paren {c_1 + d_2} + b_2 \cdot \paren {d_1 + c_2}       | c =  }} {{eqn | lo= =       | l = \paren {b_1 + a_2} \cdot c_1 + \paren {a_1 + b_2} \cdot d_1       | o = +       | r = a_2 \cdot \paren {d_1 + c_2} + b_2 \cdot \paren {c_1 + d_2}       | c = as $a_1 + b_2 = b_1 + a_2, c_1 + d_2 = d_1 + c_2$ }} {{eqn | lo= =       | l = b_1 \cdot c_1 + a_2 \cdot c_1 + a_1 \cdot b_2 + a_1 \cdot d_1       | o = +       | r = a_2 \cdot d_1 + a_2 \cdot c_2 + b_2 \cdot c_1 + b_2 \cdot d_2       | c =  }} {{eqn | lo= =       | l = \paren {a_1 \cdot d_1 + b_1 \cdot c_1 + a_2 \cdot c_2 + b_2 \cdot d_2}       | o = +       | r = \paren {a_2 \cdot c_1 + b_2 \cdot c_1 + a_2 \cdot d_1 + b_2 \cdot d_1}       | c =  }} {{end-eqn}} So we have $a_1 \cdot c_1 + b_1 \cdot d_1 + a_2 \cdot d_2 + b_2 \cdot c_2 = a_1 \cdot d_1 + b_1 \cdot c_1 + a_2 \cdot c_2 + b_2 \cdot d_2$ and so, by the definition of $\boxtimes$, we have: {{explain|The above uses cancellability of natural numbers under addition, which has been glossed over so far and needs to be made explicit.}} :$\eqclass {a_1 \cdot c_1 + b_1 \cdot d_1, a_1 \cdot d_1 + b_1 \cdot c_1} {} = \eqclass {a_2 \cdot c_2 + b_2 \cdot d_2, a_2 \cdot d_2 + b_2 \cdot c_2} {}$ So, by the definition of integer multiplication, this leads to: :$\eqclass {a_1, b_1} {} \otimes \eqclass {c_1, d_1} {} = \eqclass {a_2, b_2} {} \otimes \eqclass {c_2, d_2} {}$ Thus integer multiplication has been shown to be well-defined. {{Qed}} {{WIP|This progresses the theory for direct implementation of cross-relation on $\N \times \N$. Needs to be linked to the general approach, which is instantiated in the existing analysis of the inverse completion. The latter may be the most general approach.}}
{{begin-eqn}} {{eqn | l = \eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {}       | r = \eqclass {a, b} {} \oplus \eqclass {c_2, d_2} {}       | c =  }} {{eqn | ll= \leadsto       | l = \eqclass {b, a} {} \oplus \paren {\eqclass {a, b} {} \oplus \eqclass {c_1, d_1} {} }       | r = \eqclass {b, a} {} \oplus \paren {\eqclass {a, b} {} \oplus \eqclass {c_2, d_2} {} }       | c = adding $\eqclass {b, a} {}$ to both sides }} {{eqn | ll= \leadsto       | l = \paren {\eqclass {b, a} {} \oplus \eqclass {a, b} {} } \oplus \eqclass {c_1, d_1} {}       | r = \paren {\eqclass {b, a} {} \oplus \eqclass {a, b} {} } \oplus \eqclass {c_2, d_2} {}       | c = Integer Addition is Associative }} {{eqn | ll= \leadsto       | l = \eqclass {a + b, a + b} {} \oplus \eqclass {c_1, d_1} {}       | r = \eqclass {a + b, a + b} {} \oplus \eqclass {c_2, d_2} {}       | c = Definition of $\oplus$ }} {{eqn | ll= \leadsto       | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c = Identity for Addition of Cross-Relation Equivalence Classes on Natural Numbers }} {{end-eqn}} {{qed}} {{WIP|This progresses the theory for direct implementation of cross-relation on $\N \times \N$. Needs to be linked to the general approach, which is instantiated in the existing analysis of the inverse completion. The latter may be the most general approach.}}
Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$. Then: {{begin-eqn}} {{eqn | l = \eqclass {a_1, b_1} {}       | r = \eqclass {a_2, b_2} {}       | c = {{Defof|Operation Induced by Direct Product}} }} {{eqn | lo= \land       | l = \eqclass {c_1, d_1} {}       | r = \eqclass {c_2, d_2} {}       | c = {{Defof|Operation Induced by Direct Product}} }} {{eqn | ll= \leadstoandfrom       | l = a_1 + b_2       | r = a_2 + b_1       | c = {{Defof|Cross-Relation}} }} {{eqn | lo= \land       | l = c_1 + d_2       | r = c_2 + d_1       | c = {{Defof|Cross-Relation}} }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \paren {a_1 + c_1} + \paren {b_2 + d_2}       | r = \paren {a_1 + b_2} + \paren {c_1 + d_2}       | c = Commutativity and associativity of $+$ }} {{eqn | r = \paren {a_2 + b_1} + \paren {c_2 + d_1}       | c = from above: $a_1 + b_2 = a_2 + b_1, c_1 + d_2 = c_2 + d_1$ }} {{eqn | r = \paren {a_2 + c_2} + \paren {b_1 + d_1}       | c = Commutativity and associativity of $+$ }} {{eqn | ll= \leadsto       | l = \tuple {a_1 + c_1, b_1 + d_1}       | o = \boxtimes       | r = \tuple {a_2 + c_2, b_2 + d_2}       | c = Definition of $\boxtimes$ }} {{eqn | ll= \leadsto       | l = \paren {\tuple {a_1, b_1} \oplus \tuple {c_1, d_1} }       | o = \boxtimes       | r = \paren {\tuple {a_2, b_2} \oplus \tuple {c_2, d_2} }       | c = Definition of $\oplus$ }} {{end-eqn}} {{qed}} {{wtd|This progresses the theory for direct implementation of cross-relation on $\N \times \N$. Needs to be linked to the general approach, which is instantiated in the existing analysis of the inverse completion. The latter may be the most general approach.}}
From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation. We now need to show that: {{begin-eqn}} {{eqn | l = \tuple {x_1, y_1}       | o = \boxtimes       | r = \tuple {x_2, y_2}       | c =  }} {{eqn | lo= \land       | l = \tuple {u_1, v_1}       | o = \boxtimes       | r = \tuple {u_2, v_2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\tuple {x_1, y_1} \oplus \tuple {u_1, v_1} }       | o = \boxtimes       | r = \paren {\tuple {x_2, y_2} \oplus \tuple {u_2, v_2} }       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \tuple {x_1, y_1}       | o = \boxtimes       | r = \tuple {x_2, y_2}       | c =  }} {{eqn | lo= \land       | l = \tuple {u_1, v_1}       | o = \boxtimes       | r = \tuple {u_2, v_2}       | c = By assumption }} {{eqn | l = \paren {x_1 \circ u_1} \circ \paren {y_2 \circ v_2}       | r = \paren {x_1 \circ y_2} \circ \paren {u_1 \circ v_2}       | c = Commutativity and associativity of $\circ$ }} {{eqn | r = \paren {x_2 \circ y_1} \circ \paren {u_2 \circ v_1}       | c = By assumption }} {{eqn | r = \paren {x_2 \circ u_2} \circ \paren {y_1 \circ v_1}       | c = Commutativity and associativity of $\circ$ }} {{eqn | ll= \leadsto       | l = \tuple {x_1 \circ u_1, y_1 \circ v_1}       | o = \boxtimes       | r = \tuple {x_2 \circ u_2, y_2 \circ v_2}       | c = Definition of $\boxtimes$ }} {{eqn | ll= \leadsto       | l = \paren {\tuple {x_1, y_1} \oplus \tuple {u_1, v_1} }       | o = \boxtimes       | r = \paren {\tuple {x_2, y_2} \oplus \tuple {u_2, v_2} }       | c = Definition of $\oplus$ }} {{end-eqn}} So $\boxtimes$ is a congruence relation on $\struct {S \times C, \oplus}$. {{Qed}}
{{begin-eqn}} {{eqn | n = 1       | l = \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes       | r = \left[\!\left[{\left({y \circ b, b}\right)}\right]\!\right]_\boxtimes       | c =  }} {{eqn | l = \left({x \circ a, a}\right)       | o = \boxtimes       | r = \left({y \circ b, b}\right)       | c =  }} {{eqn | ll= \iff       | l = x \circ a \circ b       | r = y \circ b \circ a       | c = Definition of $\boxtimes$ }} {{eqn | r = y \circ a \circ b       | c = Commutativity of $\circ$ }} {{eqn | ll= \iff       | l = x       | r = y       | c = Cancellability of $a \circ b$ }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 2       | l = \left({x \circ a}\right) \circ y       | r = x \circ \left({a \circ y}\right)       | c = as $\circ$ is associative }} {{eqn | ll= \iff       | l = \left({x \circ a, y \circ a}\right)       | o = \boxtimes       | r = \left({x, y}\right)       | c = Definition of $\boxtimes$ }} {{eqn | ll= \iff       | l = \left[\!\left[{\left({x \circ a, y \circ a}\right)}\right]\!\right]_\boxtimes       | r = \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes       | c =  }} {{end-eqn}} {{Qed}}
Note that in order for $\tuple {c, c}$ and $\tuple {d, d}$ to be defined, $c$ and $d$ must be in both $S_1$ and $S_2$. Hence the restriction given: :$\forall c, d \in S_1 \cap S_2$ Then: {{begin-eqn}} {{eqn | l = \forall c, d \in S_1 \cap S_2: c \circ d       | r = d \circ c       | c = Commutativity of $\circ$ }} {{eqn | ll= \leadsto       | l = \tuple {c, c}       | o = \boxtimes       | r = \tuple {d, d}       | c = Definition of $\boxtimes$ }} {{end-eqn}} Hence the result. {{Qed}}
Because of the unique readability of $ST$, we can determine for each symbol $s$ that is part of $ST$, whether: :$s$ is part of $S$ :$s$ is part of $T$ and furthermore, precisely one of these options occurs. There are $\left\vert{S}\right\vert$ symbols in $S$, and $\left\vert{T}\right\vert$ symbols in $T$. In total, then, $ST$ is seen to consist of $\left\vert{S}\right\vert + \left\vert{T}\right\vert$ symbols. {{qed}} Category:Formal Systems d0o1ugpai8p0eyecwjkg0ghq30mujzo
{{begin-eqn}} {{eqn | l=\operatorname{Im}^{-1} \left ({\mathcal R}\right)       | r=\{s \in \operatorname{Dom} \left({\mathcal R}\right):\exists t \in \operatorname{Cdm} \left({\mathcal R}\right):(s,t)\in \mathcal R \}       | c=Definition of preimage }} {{eqn | l=\operatorname{Dom} \left ({\mathcal R}\right)       | r=\operatorname{Cdm} \left({\mathcal R}\right) = S       | c=Definition of serial relation }} {{eqn | l=\implies \operatorname{Im}^{-1} \left ({\mathcal R}\right)       | r=\{s \in S:\exists t \in S:(s,t)\in \mathcal R \} }} {{eqn | l=\forall s \in S :         | o=\exists       | r=t \in S : (s,t) \in {\mathcal R}       | c=Definition of serial relation }} {{eqn | l=\implies \forall s \in S :        | o=s       | r=\in \operatorname{Im}^{-1} \left ({\mathcal R}\right) }} {{end-eqn}} Hence the result: :$\operatorname{Im}^{-1} \left ({\mathcal R}\right) = \operatorname{Dom} \left ({\mathcal R}\right)$ {{qed}} Category:Serial Relations ib4sgq3hjvcp3hnlo04kboog05w38p1
=== Necessary Condition === Let $\mathcal R$ be a relation on $S$ which is both connected and reflexive. Let $\tuple {a, b} \in S \times S$. Suppose $a = b$. Then as $\mathcal R$ is reflexive, $\tuple {a, b} \in \mathcal R$. Suppose $a \ne b$. Then as $\mathcal R$ is connected, $\tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$. That is: :$\forall \tuple {a, b} \in S \times S: \tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$. Thus $\mathcal R$ is a total relation. {{qed|lemma}} === Sufficient Condition === Let $\mathcal R$ be a total relation on $S$. Then: :$\forall \tuple {a, b} \in S \times S: \tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$ by definition. Thus: :$\forall \tuple {a, a} \in S \times S: \tuple {a, a} \in \mathcal R$ and so $\mathcal R$ is reflexive. If $a \ne b$ the condition still holds, and so: :$\forall \tuple {a, b} \in S \times S: a \ne b \implies \tuple {a, b} \in \mathcal R \lor \tuple {b, a} \in \mathcal R$ and so $\mathcal R$ is connected. {{qed}} Category:Total Relations Category:Reflexive Relations Category:Connected Relations 19iatqoaxtgc5r1nh40ywokseick2y3
By definition of total ordering: :$\mathcal R$ is a reflexive relation on the strength of being an ordering :$\mathcal R$ is a connected relation on the strength of being a total ordering. {{qed}} Category:Total Orderings Category:Total Relations 3aq5xhgmio6wnvpnkdcsrdrap6qlg68
Using the axiomatization: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$n + 1 = 1 + n$ === Basis for the Induction === Setting $n = 1$ we have that: :$1 + 1 = 1 + 1$ and so $P \left({1}\right)$ holds trivially. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$k + 1 = 1 + k$ Then we need to show: :$\left({k + 1}\right) + 1 = 1 + \left({k + 1}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = 1 + \left({k + 1}\right)       | r = \left({1 + k}\right) + 1       | c = Axiom $C$ }} {{eqn | r = \left({k + 1}\right) + 1       | c = Induction Hypothesis }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. {{qed}} </onlyinclude>
Let $a < b$ and $b < c$. Then: {{begin-eqn}} {{eqn | l = \exists x \in \N_{> 0}: a + x       | r = b       | c =  }} {{eqn | lo= \land       | l = \exists y \in \N_{> 0}: b + y       | r = c       | c =  }} {{eqn | ll= \implies       | l = \left({a + x}\right) + y       | r = c       | c = substituting $a + x$ for $b$ }} {{eqn | ll= \implies       | l = a + \left({x + y}\right)       | r = c       | c = Natural Number Addition is Associative }} {{eqn | ll= \implies       | l = a       | o = <       | r = c       | c = as $x + y \in \N_{> 0}$ }} {{end-eqn}} {{qed}}
Let $\RR$ be a trichotomy on a set $S$. Let $x \in S$. By definition of a trichotomy, for all $a, b \in S$, either: :$a \mathrel \RR b$ :$a = b$ :$b \mathrel \RR a$ As $x = x$ it follows directly that $x \not < x$. Hence the result by definition of antireflexive relation. {{qed}} Category:Reflexive Relations Category:Trichotomies ptp4zxds19rn783zgv0p7hk2sxhdkm8
From Ordering on $1$-Based Natural Numbers is Trichotomy we have that $<$ is trichotomy. From Ordering on $1$-Based Natural Numbers is Transitive we have that $<$ is transitive. From Trichotomy is Antireflexive it follows that $<$ is antireflexive. It follows by definition that $<$ is a strict ordering. By the trichotomy law it follows that $<$ is a strict total ordering. {{qed}}
Let $\circ$ be both left cancellable and commutative on a set $S$. Then: {{begin-eqn}} {{eqn | l = a \circ c       | r = b \circ c       | c =  }} {{eqn | ll= \leadsto       | l = c \circ a       | r = c \circ b       | c = $\circ$ is Commutative }} {{eqn | ll= \leadsto       | l = a       | r = b       | c = $\circ$ is Left Cancellable }} {{end-eqn}} {{qed}}
Let $\circ$ be both right cancellable and commutative on a set $S$. Then: {{begin-eqn}} {{eqn | l = a \circ b       | r = a \circ c       | c =  }} {{eqn | ll= \leadsto       | l = b \circ a       | r = c \circ a       | c = $\circ$ is Commutative }} {{eqn | ll= \leadsto       | l = b       | r = c       | c = $\circ$ is Right Cancellable }} {{end-eqn}} {{qed}}
By Natural Number Multiplication is Commutative, proving one of the assertions suffices. From Ordering on Natural Numbers is Compatible with Multiplication it follows that: :$\forall x, y, z \in \N: x < y \iff x \times z < y \times z$ Interchanging $x$ and $y$, we obtain: :$\forall x, y, z \in \N: y < x \iff y \times z < x \times z$ From Ordering on Natural Numbers is Trichotomy, the only remaining possibility is: :$\forall x, y, z \in \N: x = y \iff x \times z = y \times z$ Hence, $\times$ is cancellable on $\N_{>0}$. {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Suppose that $\exists n \in \N_{>0}: a < n < a + 1$. Then by the definition of ordering on natural numbers: {{begin-eqn}} {{eqn | l = a + x       | r = n       | c = Definition of Ordering on Natural Numbers: $a < n$ }} {{eqn | l = n + y       | r = a + 1       | c = Definition of Ordering on Natural Numbers: $a < n$ }} {{eqn | ll= \implies       | l = \left({a + x}\right) + y       | r = a + 1       | c = substitution for $n$ }} {{eqn | ll= \implies       | l = a + \left({x + y}\right)       | r = a + 1       | c = Natural Number Addition is Associative }} {{eqn | ll= \implies       | l = x + y       | r = 1       | c = Addition on $1$-Based Natural Numbers is Cancellable }} {{end-eqn}} By Axiom $D$, either: :$y = 1$ or: :$y = t + 1$ for some $t \in \N_{>0}$ Then either: :$x + 1 = 1$ when $y = 1$ or: :$x + \left({t + 1}\right) = \left({x + t}\right) + 1 = 1$ when $y = t + 1$ Both of these conclusions violate Natural Number is Not Equal to Successor. Hence the result. {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$n \ne n + 1$ === Basis for the Induction === $P \left({1}\right)$ is the case: :$1 \ne 1 + 1$ From axiom $E$: :$E: \quad \forall a, b \in \N_{> 0}: \text{ exactly 1 of these three holds: } a = b \lor \left({\exists x \in \N_{> 0}: a + x = b}\right) \lor \left({\exists y \in \N_{> 0}: a = b + y}\right)$ Putting $a = b = 1$ this means: :$1 = 1$ or :$1 = 1 + 1$ As $1 = 1$ it follows that $1 \ne 1 + 1$ and so $P \left({1}\right)$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$k \ne k + 1$ Then we need to show: :$k + 1 \ne \left({k + 1}\right) + 1$ === Induction Step === This is our induction step: As $k \ne k + 1$  from the induction hypothesis, it follows that: :$k + 1 \ne \left({k + 1}\right) + 1$ So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{> 0}: n \ne n + 1$ {{qed}}