hcju/proofwikiqa huggingface dataset:
Let $a$ be a set. From the Axiom of Pairing the set $\set {a, a}$ may be formed. From the Axiom of Extension it follows that: :$\set {a, a} = \set a$ {{qed}}
Let $x \in S \cap T$. Then by definition of set intersection, $x \in S \land x \in T$. From Conjunction implies Disjunction of Conjunctions with Complements, it follows that: :$\paren {x \in S \land \psi} \lor \paren {x \in T \land \neg \psi}$ where $\psi$ is any arbitrary statement. Let $\psi$ be the statement $x \in R$. Thus: :$\paren {x \in S \land x \in T} \implies \paren {x \in S \land x \in R} \lor \paren {x \in T \land x \notin R}$ By definition of subset, set intersection, set union and set complement, it follows that: :$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$ {{qed}}
Let $x \in \left({R \cup S}\right) \cap \left({\overline R \cup T}\right)$. Then by definition of set intersection, set union and set complement, we have: :$\left({x \in R \lor x \in S}\right) \land \left({x \notin R \lor x \in T}\right)$ From Conjunction of Disjunctions with Complements implies Disjunction it follows that: :$x \in S \lor x \in T$ By definition of set union, it follows that: :$x \in S \cup T$ By definition of subset it follows that: :$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$ {{qed}}
{{BeginTableau|\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q}} {{Premise|1|\paren {p \lor r} \land \paren {q \lor \neg r} }} {{Simplification|2|1|p \lor r|1|1|The aim is to use Proof by Cases on this ...}} {{Assumption|3|p|Assume the first of the disjuncts ...}} {{Addition|4|3|p \lor q|3|1|... and demonstrate the conclusion}} {{Simplification|5|1|q \lor \neg r|1|2}} {{Commutation|6|1|\neg r \lor q|5|Disjunction}} {{SequentIntro|7|1|r \implies q|6|Rule of Material Implication}} {{Assumption|8|r| then assume the second of the disjuncts ...}} {{ModusPonens|9|1, 8|q|7|8}} {{Addition|10|1, 8|p \lor q|9|2|... and demonstrate the conclusion}} {{ProofByCases|11|1|p \lor q|2|3|4|8|10}} {{EndTableau}} {{qed}} Category:Conjunction Category:Disjunction 8415phdouelyojbky2zc0dg3vml37w2
By Intersection is Subset: :$\displaystyle \forall X \in P \left({S}\right): \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X \subseteq X$ From Empty Set is Element of Power Set: : $\varnothing \in P \left({S}\right)$ So: :$\displaystyle \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X \subseteq \varnothing$ From Empty Set is Subset of All Sets: :$\displaystyle \varnothing \subseteq \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X$ So by definition of set equality: :$\displaystyle \bigcap_{X \mathop \in \mathcal P \left({S}\right)} X = \varnothing$ {{qed}}
{{begin-eqn}} {{eqn | l = X       | o = \in       | r = \powerset S       | c = }} {{eqn | ll= \leadsto       | l = X       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = X       | o = \subseteq       | r = T       | c = as $S \subseteq T$: Subset Relation is Transitive }} {{eqn | l = X       | o = \in       | r = \powerset T       | c = {{Defof|Power Set}} }} {{end-eqn}} {{qed}}
By Subset of Union: :$\displaystyle \forall X \in \powerset S: X \subseteq \bigcup_{X \mathop \in \powerset S} X$ From Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$. So: :$\displaystyle S \subseteq \bigcup_{X \mathop \in \powerset S} X$ From Union is Smallest Superset: :$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$ where $\mathbb S \subseteq \powerset S$. But as $\powerset S \subseteq \powerset S$ from Set is Subset of Itself: :$\displaystyle \paren {\forall X \in \powerset S: X \subseteq S} \iff \bigcup \powerset S \subseteq S$ The {{LHS}} is no more than the definition of the power set, making it a tautological statement, and so: :$\displaystyle \bigcup \powerset S \subseteq S$ The result follows by definition of set equality. {{qed}}
=== Equality of Ordered Pairs === From Equality of Ordered Pairs, we have that: :$\set {\set a, \set {a, b} } = \set {\set c, \set {c, d} } \iff a = c, b = d$ hence showing that the Kuratowski formalization fulfils the requirement for equality. === Existence of Cartesian Product === Let $A$ and $B$ be sets. Suppose that either $A$ or $B$ is empty. Then: {{begin-eqn}} {{eqn | l = A \times B       | r = \set {\tuple {a, b} : a \in A \land b \in B}       | c = {{Defof|Cartesian Product}} }} {{eqn | r = \O }} {{end-eqn}} By the Axiom of the Empty Set $A \times B$ is a set. Now suppose that $A$ and $B$ are both non-empty. Let $a \in A$ and $b \in B$. By Singleton of Element is Subset: :$\set a \subseteq A$ and $\set b \subseteq B$ Therefore by Set Union Preserves Subsets: :$\set {a, b} \subseteq A \cup B$ $A \cup B$ is a set by Union of Small Classes is Small. Because $\set a \subseteq A \subseteq A \cup B$, it follows that: :$\set a, \set {a, b} \in \powerset {A \cup B}$ where $\powerset {A \cup B}$ is the power set of $A \cup B$. This a set by the Axiom of Powers. Thus: :$\set {\set a, \set {a, b} } \subseteq \powerset {A \cup B}$ and so by definition of the power set: :$\set {\set a, \set {a, b} } \in \powerset {\powerset {A \cup B} }$ It has been shown that: :$\set {\tuple {a, b}: a \in A \land b \in B} \subseteq \powerset {\powerset {A \cup B} }$ Applying the axiom of specification and the axiom of extension, the unique set $A \times B$ is created which consists exactly of ordered pairs $\tuple {a, b}$ such that $a \in A$ and $b \in B$. Thus it has been demonstrated that if $A$ and $B$ are non-empty, then the cartesian product $A \times B$ exists and is non-empty. Finally, in Subset of Cartesian Product it is demonstrated that every set of ordered pairs is a subset of the cartesian product of two sets. {{qed}}
Let $S$ be a set of ordered pairs. Let $x \in S$ such that $x = \left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\}$ as defined in Kuratowski Formalization of Ordered Pair. Since the elements of $S$ are sets, we can form the union $\mathbb S = \bigcup S$ of the sets in $S$. Since $x \in S$ it follows that the elements of $x$ are elements of $\mathbb S$. Since $\left\{{a, b}\right\} \in x$ it follows that $\left\{{a, b}\right\} \in \mathbb S$. Now we can form the union $\mathbb S' = \bigcup \mathbb S$ of the sets in $\mathbb S$. Since $\left\{{a, b}\right\} \in \mathbb S$ it follows that both $a$ and $b$ are elements of $\mathbb S' = \bigcup \bigcup S$. Thus from the Kuratowski Formalization of Ordered Pair we have that $S$ is a subset of some $A \times B$. We can at this stage take both $A$ and $B$ as being equal to $\bigcup \bigcup S$. Finally, the axiom of specification is applied to construct the sets: :$A = \left\{{a: \exists b: \left({a, b}\right) \in S}\right\}$ and :$B = \left\{{b: \exists a: \left({a, b}\right) \in S}\right\}$ $A$ and $B$ are seen to be the first and second projections respectively of $S$. {{qed}}
It is confirmed that, from Diagonal Relation is Equivalence, $\Delta_S$ is an equivalence relation. Let $\mathcal E$ be an arbitrary equivalence relation. By definition, $\mathcal E$ is reflexive. From Relation Contains Diagonal Relation iff Reflexive it follows that as $\Delta_S \subseteq \mathcal E$. {{qed}}
The trivial relation $\mathcal T$ on $S$ is defined as: :$\mathcal T = S \times S$ It is confirmed from Trivial Relation is Equivalence that the trivial relation is in fact an equivalence relation. Let $\mathcal E$ be an arbitrary equivalence relation on $S$. By definition of relation, $\mathcal E \subseteq S \times S$ and so (trivially) $\mathcal E \subseteq \mathcal T$. {{qed}}
It intuitively makes sense to define the length of a line segment to be the distance between the two end points, as given by the Distance Formula: :$\sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$ Similarly, it is reasonable to assume that the actual length of the curve would be approximately equal to the sum of the lengths of each of the line segments, as shown: 400px {{handwaving|The intuitive arguments need to be replaced with something more rigorous, which would probably happen on another page and be linked.}} To calculate the sum of the length of these line segments, divide $\closedint a b$ into any number of closed subintervals of the form $\closedint {x_{i - 1} } {x_i}$ where: :$a = x_0 < x_1 < \dotsb < x_{k - 1} < x_k = b$ Define: :$\Delta x_i = x_i - x_{i - 1}$ :$\Delta y_i = y_i - y_{i - 1}$ As the length of the $i$th line segment is $\sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$, the sum of all these line segments is given by: :$\displaystyle \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\Delta y_i}^2}$ {{begin-eqn}} {{eqn | r = \sum_{i \mathop = 1}^k \ \sqrt {\paren {\Delta x_i}^2 + \paren {\frac {\Delta y_i} {\Delta x_i} {\Delta x_i} }^2}       | c = multiply the second term in the radicand by $1 = \dfrac {\Delta x_i} {\Delta x_i}$ }} {{eqn | r = \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2} \Delta x_i       | c = factor $\Delta x_i$ out of the radicand }} {{end-eqn}} Thus the approximate arc length is given by the sum: :$\displaystyle s \approx \sum_{i \mathop = 1}^k \sqrt {1 + \paren {\frac {\Delta y_i} {\Delta x_i} }^2}\Delta x_i$ Recall that by hypothesis: :$f$ is continuous on $\closedint a b$ :$f$ is differentiable on $\openint a b$. Thus the Mean Value Theorem can be applied. In every open interval $\openint {x_i} {x_{i - 1} }$ there exists some $c_i$ such that: :$D_x \, \map f {c_i} = \dfrac {\Delta y_i} {\Delta x_i}$ Plugging this into the above sum we have: :$\displaystyle s \approx \sum_{i \mathop = 1}^k \ \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \Delta x_i$ By hypothesis, $D_x f$ is continuous. As Square of Real Number is Non-Negative the radicand is always positive. From Continuity of Root Function and Limit of Composite Function, $\sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$ is continuous as well. Because Continuous Real Function is Darboux Integrable, there exists a definite integral that confirms the intuitive notion that there is a value that represents the exact length of the curve as the limit of the above Riemann sum. This definite integral is: :$\displaystyle s = \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$ {{qed}}
{{begin-eqn}} {{eqn | l = s       | r = \int_a^b \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x       | c = {{Defof|Arc Length}} }} {{eqn | r = \int_a^b \sqrt {\paren {\frac {\frac {\d x} {\d t} } {\frac {\d x} {\d t} } }^2 + \paren {\frac {\frac {\d y}{\d t} } {\frac {\d x} {\d t} } }^2} \rd x       | c = because $\paren {\dfrac {\frac {\d x} {\d t} } {\frac {\d x}{\d t} } }^2 = 1$, and from corollary to chain rule }} {{eqn | r = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \paren {\frac 1 {\frac {\d x} {\d t} } } \rd x       | c = factoring $\dfrac {\d x}{\d t}$ out of the radicand. No absolute value is needed as length cannot be negative. }} {{eqn | r = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \frac {\d t} {\d x} \rd x       | c = Derivative of Inverse Function }} {{eqn | r = \int_a^b \sqrt {\paren {\frac {\d x} {\d t} }^2 + \paren {\frac {\d y} {\d t} }^2} \rd t       | c = Integration by Substitution }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $y = \map f x$. From the definition of the preimage of an element: :$\map {f^{-1} } y = \set {x \in S: \tuple {y, x} \in f}$ Thus: :$x \in \map {f^{-1} } y$ By definition of injection, $\map {f^{-1} } y$ is a singleton: :$\map {f^{-1} } y = \set x$ which can be expressed as: :$\map {f^{-1} } y = x$ {{qed|lemma}} === Sufficient Condition === Let $\map {f^{-1} } y = x$. Thus by definition of direct image mapping: :$\map {f^\gets} {\set y} = \set x$ Then: :$\map {f^\to} {\map {f^\gets} {\set y} } = \map {f^\to} {\set x}$ So from the corollary to Image of Preimage under Mapping: :$\set y = \map {f^\to} {\set x}$ and so by definition of direct image mapping: :$y = \map f x$ {{qed}}
From definition, a set $X$ is an infinite successor set {{iff}} both the following holds: :$\O \in X$ :$x \in X \implies x^+ \in X$ For all $S \in \mathbb S$, $S$ is an infinite successor set. From definition of an infinite successor set, $\O \in S$. By definition of set intersection, $\O \in \bigcap \mathbb S$. Now suppose $x \in \bigcap \mathbb S$. By definition of set intersection: :$\forall S \in \mathbb S: x \in S$ Hence from definition of an infinite successor set, $x^+ \in S$. By definition of set intersection, $x^+ \in \bigcap \mathbb S$. Therefore $\bigcap \mathbb S$ is an infinite successor set. {{qed}}
From the Axiom of Infinity, there is a set $S$ such that: :$\O \in S$ :$x \in S \implies x^+ \in S$ where $\O$ denotes the empty set and $x^+$ is the successor set of $x$. That is, there exists an infinite successor set. Next, by the Axiom of Specification, the minimal infinite successor set $\omega$: :$\omega := \displaystyle \bigcap \set {S' \subseteq S: \text{$S'$ is an infinite successor set} }$ exists. It remains to be shown that if $T$ is an infinite successor set, then $\omega \subseteq T$. By Intersection of Infinite Successor Sets, $S \cap T$ is an infinite successor set. Moreover, by Intersection is Subset, $S \cap T$, so that: :$S \cap T \in \set {S' \subseteq S: \text{$S'$ is an infinite successor set} }$ from which we conclude by definition of $\omega$ and Intersection is Subset: General Result that: :$\omega \subseteq S \cap T \subseteq T$ as desired. {{qed}}
We need to check that all of Peano's axioms hold for $\left({\omega, \cdot^+, \varnothing}\right)$. Suppose first that for $m, n \in \omega$, we have $m^+ = n^+$. Since $n \in n^+$ it follows that $n \in m^+$. Hence, either $n \in m$ or $n = m$. Similarly, either $m \in n$ or $m = n$. Now if $n \ne m$, both $m \in n$ and $n \in m$. By Element of Minimal Infinite Successor Set is Transitive Set, it follows that $n \subseteq m$. As $m \in n$, this contradicts Finite Ordinal is not Subset of one of its Elements. Hence it must be that $n = m$, and Axiom $(P3)$ holds. Next, since $n \in n^+$ for all $n \in \omega$, it follows that $n^+ \ne \varnothing$. Hence, Axiom $(P4)$ holds as well. Finally, let $S \subseteq \omega$ satisfy: :$\varnothing \in S$ :$\forall n \in S: n^+ \in S$ Then by definition, $S$ is an infinite successor set. Therefore, by definition of $\omega$ as the minimal infinite successor set: :$\omega \subseteq S$ Consequently $S = \omega$ by the definition of set equality. Thus Axiom $(P5)$ is seen to hold. That is, $\left({\omega, \cdot^+, \varnothing}\right)$ is a Peano structure. {{qed}}
=== Base $e$ === :$\displaystyle \int \ln x \ \mathrm d x = \int 1 \cdot \ln x \ \mathrm d x$ From Integration by Parts: :$\displaystyle \int fg' \, \mathrm dx = fg - \int f'g \, \mathrm dx$ Here: {{begin-eqn}} {{eqn | l = f       | r = \ln x }} {{eqn | ll = \implies       | l = f'       | r = \frac 1 x       | c = Derivative of Natural Logarithm Function }} {{eqn | l = g'       | r = 1 }} {{eqn | ll = \implies       | l = g       | r = x       | c = Integral of Constant }} {{end-eqn}} So: {{begin-eqn}} {{eqn| l = \int \ln x \ \mathrm d x      | r = x \cdot \ln x -\int x \cdot \frac 1 x \ \mathrm dx }} {{eqn | r = x \ln x - x + C }} {{end-eqn}} {{qed}} === Base $a$ === Follows from Linear Combination of Integrals and Linear Operator on General Logarithm. {{qed}} Category:Integral Calculus fzp9dlrzzftz592g530wc8klw6j0asy
Let $S$ be the set of all those finite ordinals $n$ which are not a subset of any of its elements. That is: :$n \in S \iff n \in \omega \land \forall x \in n: n \nsubseteq x$ We know that $0 = \varnothing$ is not a subset of any of its elements, as $\varnothing$ by definition has no elements. So $0 \in S$. Now suppose $n \in S$. From Set is Subset of Itself: :$n \subseteq n$ But as $n \in S$ it follows by definition of $S$ that: :$n \notin n$ By definition of the successor of $n$, it follows that: :$n^+ \nsubseteq n$ Now from Subset Relation is Transitive: :$n^+ \subseteq x \implies n \subseteq x$ But since $n \in S$ it follows that: :$x \notin n$ So: :$n^+ \nsubseteq n$ and: :$\forall x \in n: n^+ \nsubseteq x$ So $n^+$ is not a subset of any of its elements. That is: :$n^+ \in S$ So by the Principle of Mathematical Induction: :$S = \omega$ Hence the result. {{qed}}
Let $S \subseteq$ be the set of all transitive elements of $\omega$. That is: :$n \in S \iff n \in \omega \land \forall x \in n: x \subseteq n$ It is vacuously true that $0 \in S$, as there are no $x \in 0$. Now suppose $n \in S$. If $x \in n^+$ then either $x \in n$ or $x = n$. In the first case: :$x \subseteq n$ as $n \in S$ and so: :$x \subseteq n^+$ In the second case, by definition of successor set: :$x \subseteq n^+$ By Principle of Mathematical Induction for Minimal Infinite Successor Set it follows that: :$S = \omega$ Hence the result. {{qed}}
Proof by induction: From the definition of addition, we have that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m + 0       | r = m       | c =  }} {{eqn | l = m + n^+       | r = \left({m + n}\right)^+       | c =  }} {{end-eqn}} For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$0 + n = n = n + 0$ === Basis for the Induction === By definition, we have: :$0 + 0 = 0 = 0 + 0$ Thus $P \left({0}\right)$ is seen to be true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true. So this is our induction hypothesis $P \left({k}\right)$: :$0 + k = k = k + 0$ Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$: :$0 + k^+ = k^+ = k^+ + 0$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = 0 + k^+       | r = \left({0 + k}\right)^+       | c = Definition of Addition }} {{eqn | r = \left({k + 0}\right)^+       | c = from the induction hypothesis }} {{eqn | r = k^+       | c = Definition of Addition: $k + 0 = k$ }} {{end-eqn}} By definition: :$k^+ + 0 = k^+$ So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: 0 + n = n = n + 0$ {{qed}}
Proof by induction: From the definition of addition, we have that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m + 0       | r = m       | c =  }} {{eqn | l = m + n^+       | r = \left({m + n}\right)^+       | c =  }} {{end-eqn}} For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :$\forall m \in \N: m^+ + n = \left({m + n}\right)^+$ === Basis for the Induction === By definition, we have: {{begin-eqn}} {{eqn | ll= \forall m \in \N:       | l = m^+ + 0       | r = m^+       | c = by definition }} {{eqn | r = \left({m + 0}\right)^+       | c = by definition: $m = m + 0$ }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to be true. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true. So this is our induction hypothesis $P \left({k}\right)$: :$\forall m \in \N: m^+ + k = \left({m + k}\right)^+$ Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$: :$\forall m \in \N: m^+ + k^+ = \left({m + k^+}\right)^+$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = m^+ + k^+       | r = \left({m^+ + k}\right)^+       | c = by definition }} {{eqn | r = \left({\left({m + k}\right)^+}\right)^+       | c = from the induction hypothesis }} {{eqn | r = \left({m + k^+}\right)^+       | c = by definition }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n \in \N: m^+ + n = \left({m + n}\right)^+$ {{qed}}
Using the following axioms: {{:Axiom:Axiomatization of 1-Based Natural Numbers}} Proof by induction: From Axiom:Axiomatization of 1-Based Natural Numbers, we have by definition that: {{begin-eqn}} {{eqn | ll= \forall m, n \in \N:       | l = m + 0       | r = m }} {{eqn | l = \left({m + n}\right)^+       | r = m + n^+ }} {{end-eqn}} For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\forall m \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$ === Basis for the Induction === When $n = 1$, we have: :$\left({m + 1}\right) + 1 = \left({m + 1}\right) + 1$ which holds trivially. Thus $P \left({1}\right)$ is seen to be true. 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 $P \left({k}\right)$: :$\forall m \in \N: \left({m + 1}\right) + k = \left({m + k}\right) + 1$ Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$: :$\forall m \in \N: \left({m + 1}\right) + \left({k + 1}\right) = \left({m + \left({k + 1}\right)}\right) + 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \left({m + \left({k + 1}\right)}\right) + 1       | r = \left({\left({m + k}\right) + 1}\right) + 1       | c = Natural Number Addition is Associative/Proof 3 }} {{eqn | r = \left({\left({m + 1}\right) + k}\right) + 1       | c = from the induction hypothesis }} {{eqn | r = \left({m + 1}\right) + \left({k + 1}\right)       | c = Natural Number Addition is Associative/Proof 3 }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$ {{qed}}