hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = -\paren {x - y}       | r = -\paren {x + \paren {-y} }       | c = {{Defof|Real Subtraction}} }} {{eqn | r = -x - \paren {-y}       | c = Negative of Sum of Real Numbers }} {{eqn | r = -x + \paren {-\paren {-y} }       | c = {{Defof|Real Subtraction}} }} {{eqn | r = -x + y       | c = Negative of Negative Real Number }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x \times y       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \frac 1 x \times \paren {x \times y}       | r = \frac 1 x \times x       | c = as long as $x \ne 0$ }} {{eqn | ll= \leadsto       | l = \paren {\frac 1 x \times x} \times y       | r = \frac 1 x \times x       | c = Real Number Axioms: $\R \text M 1$: Associativity }} {{eqn | ll= \leadsto       | l = 1 \times y       | r = 1       | c = Real Number Axioms: $\R \text M 4$: Inverses }} {{eqn | ll= \leadsto       | l = y       | r = 1       | c = Real Multiplication Identity is One }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \ne 0:       | l = \frac x x       | r = x \times \frac 1 x       | c = {{Defof|Real Division}} }} {{eqn | r = 1       | c = Real Number Axioms: $\R M 4$: Inverses }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac x 1       | r = x \times \frac 1 1       | c = {{Defof|Real Division}} }} {{eqn | r = x \times 1       | c = Real Number Divided by Itself }} {{eqn | r = x       | c = Real Number Axioms: $\R \text M 4$: Identity }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {x \times y} \times \paren {x \times y}       | r = 1       | c = Real Number Axioms: $\R \text M 4$: Inverses }} {{eqn | ll= \leadsto       | l = \frac 1 {x \times y} \times \paren {x \times y} \times \frac 1 y       | r = 1 \times \frac 1 y       | c = as $y \ne 0$ }} {{eqn | ll= \leadsto       | l = \paren {\frac 1 {x \times y} \times x} \times \paren {y \times \frac 1 y}       | r = 1 \times \frac 1 y       | c = Real Number Axioms: $\R \text M 1$: Associativity }} {{eqn | ll= \leadsto       | l = \paren {\frac 1 {x \times y} \times x} \times 1       | r = 1 \times \frac 1 y       | c = Real Number Axioms: $\R \text M 4$: Inverse }} {{eqn | ll= \leadsto       | l = \frac 1 {x \times y} \times x       | r = \frac 1 y       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{eqn | ll= \leadsto       | l = \paren {\frac 1 {x \times y} \times x} \times \frac 1 x       | r = \frac 1 y \times \frac 1 x       | c = as $x \ne 0$ }} {{eqn | ll= \leadsto       | l = \frac 1 {x \times y} \times \paren {x \times \frac 1 x}       | r = \frac 1 y \times \frac 1 x       | c = Real Number Axioms: $\R \text M 1$: Associativity }} {{eqn | ll= \leadsto       | l = \frac 1 {x \times y} \times 1       | r = \frac 1 y \times \frac 1 x       | c = Real Number Axioms: $\R \text M 4$: Inverse }} {{eqn | ll= \leadsto       | l = \frac 1 {x \times y}       | r = \frac 1 y \times \frac 1 x       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{eqn | ll= \leadsto       | l = \frac 1 {x \times y}       | r = \frac 1 x \times \frac 1 y       | c = Real Number Axioms: $\R \text M 2$: Commutativity }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac x y \times \frac w z       | r = x \times \frac 1 y \times w \times \frac 1 z       | c = {{Defof|Real Division}} }} {{eqn | r = x \times w \times \frac 1 y \times \frac 1 z       | c = Real Number Axioms: $\R \text M 2$: Commutativity }} {{eqn | r = \paren {x \times w} \times \paren {\frac 1 y \times \frac 1 z}       | c = Real Number Axioms: $\R \text M 1$: Associativity }} {{eqn | r = \paren {x \times w} \times \frac 1 {y \times z}       | c = Product of Reciprocals of Real Numbers }} {{eqn | r = \dfrac {x \times w} {y \times z}       | c = {{Defof|Real Division}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac x y + \frac w z       | r = \paren {x \times \frac 1 y} + \paren {w \times \frac 1 z}       | c = {{Defof|Real Division}} }} {{eqn | r = \paren {x \times \frac 1 y \times 1} + \paren {1 \times w \times \frac 1 z}       | c = Real Number Axioms: $\R M 3$: Identity }} {{eqn | r = \paren {x \times \frac 1 y \times z \times \frac 1 z} + \paren {y \times \frac 1 y \times w \times \frac 1 z}       | c = Real Number Axioms: $\R M 4$: Inverses }} {{eqn | r = \paren {x \times z \times \frac 1 y \times \frac 1 z} + \paren {y \times w \times \frac 1 y \times \frac 1 z}       | c = Real Number Axioms: $\R M 2$: Commutativity }} {{eqn | r = \paren {\paren {x \times z} \times \paren {\frac 1 y \times \frac 1 z} } + \paren {\paren {y \times w} \times \paren {\frac 1 y \times \frac 1 z} }       | c = Real Number Axioms: $\R M 1$: Associativity }} {{eqn | r = \paren {\paren {x \times z} + \paren {y \times w} } \times \paren {\frac 1 y \times \frac 1 z}       | c = Real Number Axioms: $\R D$: Distributivity }} {{eqn | r = \paren {\paren {x \times z} + \paren {y \times w} } \times \frac 1 {y z}       | c = Product of Reciprocals of Real Numbers }} {{eqn | r = \frac {\paren {x \times z} + \paren {y \times w} } {y \times z}       | c = {{Defof|Real Division}} }} {{end-eqn}} {{qed}}
{{AimForCont}} that: :$\exists x \in \R_{\ne 0}: \dfrac 1 x = 0$ From Real Zero is Zero Element :$\dfrac 1 x \times x = 0$ But from Real Number Axioms: $\R \text M 4$: Inverses: :$\dfrac 1 x \times x = 1$ The result follows by Proof by Contradiction. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 {x / y}       | r = \frac 1 {x \times \dfrac 1 y}       | c = {{Defof|Real Division}} }} {{eqn | r = 1 \times \frac 1 {x \times \dfrac 1 y}       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{eqn | r = \paren {y \times \frac 1 y} \times \frac 1 {x \times \dfrac 1 y}       | c = Real Number Axioms: $\R \text M 4$: Inverses }} {{eqn | r = y \times \paren {\frac 1 y \times \frac 1 {x \times \dfrac 1 y} }       | c = Real Number Axioms: $\R \text M 1$: Associativity }} {{eqn | r = y \times \frac 1 {y \times \paren {x \times \dfrac 1 y} }       | c = Product of Reciprocals of Real Numbers }} {{eqn | r = y \times \frac 1 {x \times \paren {y \times \dfrac 1 y} }       | c = Real Number Axioms: $\R \text M 1$: Associativity and $\R \text M 2$: Commutativity }} {{eqn | r = y \times \frac 1 {x \times 1}       | c = Real Number Axioms: $\R \text M 4$: Inverses }} {{eqn | r = y \times \frac 1 x       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{eqn | r = \frac y x       | c = {{Defof|Real Division}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {x / y} {w / z}       | r = \frac x y \times \frac 1 {w / z}       | c = {{Defof|Real Division}} }} {{eqn | r = \frac x y \times \frac z w       | c = Reciprocal of Quotient of Real Numbers }} {{eqn | r = \dfrac {x \times z} {y \times w}       | c = Product of Quotients of Real Numbers }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {a \times x} y       | r = \paren {a \times x} \times \frac 1 y       | c = {{Defof|Real Division}} }} {{eqn | r = a \times \paren {x \times \frac 1 y}       | c = Real Number Axioms: $\R \text M 1$: Associativity }} {{eqn | r = a \times \frac x y       | c = {{Defof|Real Division}} }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $E$ be a neighborhood of $x$ in $\struct {S, \tau_S}$.  By the definition of neighborhood: :$\exists U \in \tau_S : x \in U \subseteq E$ Now, by the definition of the subspace topology: :$\exists V \in \tau: U = V \cap S$ Take $D := V \cup E$. We have that: :$V \subseteq D$ and: :$V \in \tau$ Thus $D$ is a neighborhood of $x$ in $X$. Therefore it holds true that: :$D \cap S = \paren {V \cap S} \cup \paren {E \cap S} = U \cup E = E$ {{qed|lemma}} === Sufficient Condition === Let $\exists D \subseteq X$ such that $D$ is a neighborhood of $x$ in $\struct {X, \tau}$. Let $E = D \cap S$ Then, by the definition of neighborhood: :$\exists V \in \tau: x \in V \subseteq D$ By the definition of the subspace topology: :$V \cap S \in \tau_S$ As a consequence: :$E = D \cap S \supseteq V \cap S \in \tau_S$ That is: :$E$ is a neighborhood of $x$ in $\struct {S, \tau_S}$. {{qed}}
{{begin-eqn}} {{eqn | l = \frac {-x} y       | r = \frac {\paren {-1} \times x} y       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = \paren {-1} \times \frac x y       | c = Product of Real Number with Quotient }} {{eqn | r = - \frac x y       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = \paren {-1} \times \frac x y       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = \paren {-1} \times 1 \times \frac x y       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{eqn | r = \paren {-1} \times \frac {-1} {-1} \times \frac x y       | c = Real Number Divided by Itself }} {{eqn | r = \frac {-1 \times -1} {-1} \times \frac x y       | c = Product of Real Number with Quotient }} {{eqn | r = \frac {-\paren {-1} } {-1} \times \frac x y       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = \frac 1 {-1} \times \frac x y       | c = Negative of Negative Real Number }} {{eqn | r = \frac {1 \times x} {\paren {-1} \times y}       | c = Product of Quotients of Real Numbers }} {{eqn | r = \frac {1 \times x} {- y}       | c = Multiplication by Negative Real Number: Corollary }} {{eqn | r = \frac x {- y}       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{end-eqn}} {{qed}}
By Compact Space is Weakly Locally Compact, $T$ is weakly locally compact. Thus $T$ is a locally compact Hausdorff space. {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = >       | r = 0       | c = }} {{eqn | lo= \land       | l = y       | o = >       | r = 0       | c = Real Number Ordering is Compatible with Addition }} {{eqn | ll= \leadsto       | l = x + y       | o = >       | r = 0 + 0       | c = Real Number Inequalities can be Added }} {{eqn | r = 0       | c = Real Number Axioms: $\R \text A 3$: Identity }} {{end-eqn}} {{qed}}
Let $y > 0$. From Real Number Axioms: $\R \text O 2$: compatible with multiplication: :$x > z \implies x \times y > z \times y$ Thus setting $z = 0$: :$x > 0 \implies x \times y > 0 \times y$ But from Real Zero is Zero Element: :$0 \times y = 0$ Hence the result: :$x, y > 0 \implies x \times y > 0$ {{qed}}
Let $x > 0$. {{begin-eqn}} {{eqn | l = x       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x + \paren {-x}       | o = >       | r = 0 + \paren {-x}       | c = Real Number Axioms: $\R \text O 2$: compatibility with addition }} {{eqn | ll= \leadsto       | l = 0       | o = >       | r = 0 + \paren {-x}       | c = Real Number Axioms: $\R \text A 4$: Inverses }} {{eqn | ll= \leadsto       | l = 0       | o = >       | r = \paren {-x}       | c = Real Number Axioms: $\R \text A 3$: Identity }} {{end-eqn}} {{qed|lemma}} Let $\paren {-x} < 0$. {{begin-eqn}} {{eqn | l = \paren {-x}       | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x + \paren {-x}       | o = <       | r = x + 0       | c = Real Number Axioms: $\R \text O 2$: compatibility with addition }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = x + 0       | c = Real Number Axioms: $\R \text A 4$: Inverses }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = x       | c = Real Number Axioms: $\R \text A 3$: Identity }} {{end-eqn}} {{qed}}
Let $x > y$. {{begin-eqn}} {{eqn | l = x       | o = >       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x + \paren {-x}       | o = >       | r = y + \paren {-x}       | c = Real Number Axioms: $\R \text O 1$: compatibility with addition }} {{eqn | ll= \leadsto       | l = 0       | o = >       | r = y + \paren {-x}       | c = Real Number Axioms: $\R \text A 4$: Inverses }} {{eqn | ll= \leadsto       | l = 0 + \paren {-y}       | o = >       | r = y + \paren {-x} + \paren {-y}       | c = Real Number Axioms: $\R \text O 1$: compatibility with addition }} {{eqn | ll= \leadsto       | l = 0 + \paren {-y}       | o = >       | r = \paren {y + \paren {-y} } + \paren {-x}       | c = Real Number Axioms: $\R \text A 1$: Associativity and $\R \text A 2$: Commuativity }} {{eqn | ll= \leadsto       | l = 0 + \paren {-y}       | o = >       | r = 0 + \paren {-x}       | c = Real Number Axioms: $\R \text A 4$: Inverses }} {{eqn | ll= \leadsto       | l = \paren {-y}       | o = >       | r = \paren {-x}       | c = Real Number Axioms: $\R \text A 3$: Identity }} {{eqn | ll= \leadsto       | l = \paren {-x}       | o = <       | r = \paren {-y}       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} {{qed|lemma}} Let $\paren {-x} < \paren {-y}$. {{begin-eqn}} {{eqn | l = \paren {-x}       | o = <       | r = \paren {-y}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {-x} + x       | o = <       | r = \paren {-y} + x       | c = Real Number Axioms: $\R \text O 1$: compatibility with addition }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \paren {-y} + x       | c = Real Number Axioms: $\R \text A 4$: Inverses }} {{eqn | ll= \leadsto       | l = 0 + y       | o = <       | r = \paren {-y} + x + y       | c = Real Number Axioms: $\R \text O 1$: compatibility with addition }} {{eqn | ll= \leadsto       | l = 0 + y       | o = <       | r = \paren {\paren {-y} + y} + x       | c = Real Number Axioms: $\R \text A 1$: Associativity and $\R \text A 2$: Commuativity }} {{eqn | ll= \leadsto       | l = 0 + y       | o = <       | r = 0 + x       | c = Real Number Axioms: $\R \text A 4$: Inverses }} {{eqn | ll= \leadsto       | l = y       | o = <       | r = x       | c = Real Number Axioms: $\R \text A 3$: Identity }} {{eqn | ll= \leadsto       | l = x       | o = >       | r = y       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = x       | o = <       | r = y }} {{eqn | ll= \leadstoandfrom       | l = y - x       | o = >       | r = 0       | c = Inequality iff Difference is Positive }} {{eqn | ll= \leadstoandfrom       | lr = -x + y       | o = >       | r = 0       | c = Real Number Axioms: $\R \text A 2 $: Commutativity of Addition }} {{eqn | ll= \leadstoandfrom       | l = -x + -\paren {-y}       | o = >       | r = 0       | c = Negative of Negative Real Number }} {{eqn | ll= \leadstoandfrom       | l = -x - \paren {-y}       | o = >       | r = 0       | c = {{Defof|Real Subtraction}} }} {{eqn | ll= \leadstoandfrom       | l = -y       | o = <       | r = -x       | c = Inequality iff Difference is Positive }} {{end-eqn}} {{qed}}
There are two cases to consider: :$(1): \quad x > 0$ :$(2): \quad x < 0$ Let $x > 0$. Then: {{begin-eqn}} {{eqn | l = x \times x       | o = >       | r = 0       | c = Product of Strictly Positive Real Numbers is Strictly Positive }} {{end-eqn}} Let $x < 0$. Then: {{begin-eqn}} {{eqn | l = x       | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x \times x       | o = >       | r = x \times 0       | c = Real Number Ordering is Compatible with Multiplication: Negative Factor }} {{eqn | r = 0       | c = Real Zero is Zero Element }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 0       | o = <       | r = 1       | c = Real Zero is Less than Real One }} {{eqn | ll= \leadsto       | l = -0       | o = >       | r = -1       | c = Order of Real Numbers is Dual of Order of their Negatives }} {{eqn | ll= \leadsto       | l = 0       | o = >       | r = -1       | c = Negative of Real Zero equals Zero }} {{eqn | ll= \leadsto       | l = -1       | o = <       | r = 0       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1 \times 1       | o = >       | r = 0       | c = Square of Non-Zero Real Number is Strictly Positive }} {{eqn | ll= \leadsto       | l = 1       | o = >       | r = 0       | c = Real Number Axioms: $\R \text M 3$: Identity }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = 1       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let $x \times y > 0$. {{AimForCont}} either $x = 0$ or $y = 0$. Then from Real Zero is Zero Element: :$x \times y = 0$ Therefore by Proof by Contradiction: :$y \ne 0$ and $x \ne 0$ {{qed|lemma}} Let $x > 0$. {{AimForCont}} $y < 0$. {{begin-eqn}} {{eqn | l = x       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x \times y       | o = <       | r = 0 \times 0       | c = Real Number Ordering is Compatible with Multiplication: Negative Factor }} {{eqn | ll= \leadsto       | l = x \times y       | o = <       | r = 0       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} But by hypothesis, $x \times y > 0$. Therefore by Proof by Contradiction: :$y > 0$ {{qed|lemma}} Let $x < 0$. {{AimForCont}} $y > 0$. {{begin-eqn}} {{eqn | l = x       | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x \times y       | o = <       | r = 0 \times 0       | c = Real Number Axioms: $\R \text O 2$: compatibility with multiplication }} {{eqn | ll= \leadsto       | l = x \times y       | o = <       | r = 0       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} But by hypothesis, $x \times y > 0$. Therefore by Proof by Contradiction: :$y > 0$ {{qed|lemma}} Thus: :$x \times y > 0 \implies \paren {x > 0 \land y > 0} \lor \paren {x < 0 \land y < 0}$ {{qed|lemma}} === Necesssary Condition === Let $x > 0$ and $y > 0$. Then from Product of Strictly Positive Real Numbers is Strictly Positive: :$x \times y > 0$ {{qed|lemma}} Let $x < 0$ and $y < 0$. {{begin-eqn}} {{eqn | l = x       | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x \times y       | o = >       | r = 0 \times 0       | c = Real Number Ordering is Compatible with Multiplication: Negative Factor }} {{eqn | ll= \leadsto       | l = x \times y       | o = >       | r = 0       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} {{qed|lemma}} Thus if either $x, y > 0$ or $x, y < 0$: :$x \times y > 0$ {{qed}}
Let $x > 0$. {{AimForCont}} $\dfrac 1 x < 0$. Then: {{begin-eqn}} {{eqn | l = x       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x \times \dfrac 1 x       | o = <       | r = 0 \times 0       | c = Real Number Ordering is Compatible with Multiplication: Negative Factor }} {{eqn | ll= \leadsto       | l = 1       | o = <       | r = 0       | c = Real Number Axioms: $\R \text M 4$: Inverse }} {{end-eqn}} But from Real Zero is Less than Real One: :$1 > 0$ Therefore by Proof by Contradiction: :$\dfrac 1 x > 0$ {{qed}}
From Reciprocal of Strictly Positive Real Number is Strictly Positive: :$(1): \quad x > 0 \implies \dfrac 1 x > 0$ :$(2): \quad y > 0 \implies \dfrac 1 y > 0$ Then: {{begin-eqn}} {{eqn | l = x       | o = >       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x \times \frac 1 x       | o = >       | r = y \times \frac 1 x       | c = Real Number Axioms: $\R O2$: compatibility with multiplication and from $(1)$ }} {{eqn | ll= \leadsto       | l = x \times \frac 1 x \times \frac 1 y       | o = >       | r = y \times \frac 1 x \times \frac 1 y       | c = Real Number Axioms: $\R O2$: compatibility with multiplication and from $(2)$ }} {{eqn | ll= \leadsto       | l = \paren {x \times \frac 1 x} \times \frac 1 y       | o = >       | r = \paren {y \times \frac 1 y} \times \frac 1 x       | c = Real Number Axioms: $\R M1$ (Associativity) and $\R M2$ (Commutativity) }} {{eqn | ll= \leadsto       | l = 1 \times \frac 1 y       | o = >       | r = 1 \times \frac 1 x       | c = Real Number Axioms: $\R M4$ (Inverses) }} {{eqn | ll= \leadsto       | l = \frac 1 y       | o = >       | r = \frac 1 x       | c = Real Number Axioms: $\R M3$ (Identity) }} {{eqn | ll= \leadsto       | l = \frac 1 x       | o = <       | r = \frac 1 y       | c = {{Defof|Dual Ordering}} }} {{end-eqn}} {{qed}}
First note that: {{begin-eqn}} {{eqn | l = 0       | o = <       | r = 1       | c = Real Zero is Less than Real One }} {{eqn | ll= \leadsto       | l = 0 + 0       | o = <       | r = 1 + 1       | c = Real Number Inequalities can be Added }} {{eqn | ll= \leadsto       | l = 0       | o = <       | r = \frac 1 {1 + 1}       | c = Reciprocal of Strictly Positive Real Number is Strictly Positive }} {{eqn | n = 1       | ll= \leadsto       | l = 0       | o = <       | r = \frac 1 2       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x       | o = <       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x + x       | o = <       | r = x + y       | c = Real Number Axioms: $\R O1$: compatibility with addition }} {{eqn | ll= \leadsto       | l = \paren {x + x} \times \frac 1 2       | o = <       | r = \paren {x + y} \times \frac 1 2       | c = Real Number Axioms: $\R O2$: compatibility with multiplication and from $(1)$ }} {{eqn | ll= \leadsto       | l = x       | o = <       | r = \frac {x + y} 2       | c = {{Defof|Real Division}} }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = x       | o = <       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = x + y       | o = <       | r = y + y       | c = Real Number Axioms: $\R O1$: compatibility with addition }} {{eqn | ll= \leadsto       | l = \paren {x + y} \times \frac 1 2       | o = <       | r = \paren {y + y} \times \frac 1 2       | c = Real Number Axioms: $\R O2$: compatibility with multiplication and from $(1)$ }} {{eqn | ll= \leadsto       | l = \frac {x + y} 2       | o = <       | r = y       | c = {{Defof|Real Division}} }} {{end-eqn}} {{qed}}
By definition of inductive set: :$\forall S \in \AA: 1 \in S$ Thus by definition of set intersection: :$\displaystyle 1 \in \bigcap_{S \mathop \in \AA} S$ Also by definition of inductive set: :$\forall S \in \AA: x \in S \implies x + 1 \in S$ So: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcap_{S \mathop \in \AA} S       | c =  }} {{eqn | ll= \leadsto       | lo= \forall S \in \AA:       | l = x       | o = \in       | r = S       | c = {{Defof|Intersection of Set of Sets}} }} {{eqn | ll= \leadsto       | lo= \forall S \in \AA:       | l = x + 1       | o = \in       | r = S       | c = {{Defof|Inductive Set as Subset of Real Numbers}} }} {{eqn | ll= \leadsto       | l = x + 1       | o = \in       | r = \bigcap_{S \mathop \in \AA} S       | c = {{Defof|Intersection of Set of Sets}} }} {{end-eqn}} Hence the result, by definition of inductive set. {{qed}}
The proof will proceed by the Principle of Mathematical Induction on $\N_{>0}$. Let $S$ be the set defined as: :$S := \left\{ {n \in \N:}\right.$ Every non-empty subset of $\N^*_n$ has a greatest element$\left.{}\right\}$ === Basis for the Induction === We have that: : $\N^*_1 = \set 1$ The only non-empty subset  of $\set 1$ is $\set 1$ itself. This has a greatest element $1$. So $1 \in S$. This is our basis for the induction. === Induction Hypothesis === It is to be shown that, if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$. This is the induction hypothesis: :Every non-empty subset of $\N^*_k$ has a greatest element. It is to be demonstrated that it follows that: :Every non-empty subset of $\N^*_{k + 1}$ has a greatest element. === Induction Step === This is our induction step: Let $T \subseteq \N^*_{k + 1}$. Either $k + 1 \in T$ or $k + 1 \notin T$. If $k + 1 \notin T$ then $T \subseteq \N^*_k$. In that case, then from the induction hypothesis: : $T$ has a greatest element. If $k + 1 \in T$ then: :$\forall x \in T: x \le k + 1$ and so $k + 1$ is the greatest element of $T$. So in both cases $T$ has a greatest element and so $k + 1 \in S$. So $k \in S \implies k + 1 \in S$ and the result follows by the Principle of Mathematical Induction: :$\forall n \in \N$: every non-empty subset of $\N^*_n$ has a greatest element. {{qed}}
Let: :$S = \set {x \in \N_{>0}: x > 1}$ Then $S \subseteq \N_{>0}$. {{AimForCont}} $S$ has a greatest element. Let the greatest element of $S$ be $k$. But $\N_{>0}$ is an inductive set. Therefore $k + 1 \in \N_{>0}$. By definition of $S$: :$k + 1 \in S$ Therefore $k$ cannot be the greatest element of $S$. By Proof by Contradiction it follows that $S$ has no greatest element. It is not possible to apply Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element, because $S$ is not a subset of an initial segment of $\N_{>0}$. {{qed}}