hcju/proofwikiqa huggingface dataset:
We will show that we can choose an infinite sequence of nodes $t_0, t_1, t_2, \ldots$ of $T$ such that: * $t_0$ is the root node; * $t_{n+1}$ is a child of $t_n$; * Each $t_n$ has infinitely many descendants. Then the sequence $t_0, t_1, t_2, \ldots$ is such a branch of infinite length. Take the root node $t_0$. By definition, it has a finite number of children. Suppose that all of these childen had a finite number of descendants. Then that would mean that $t_0$ had a finite number of descendants, and that would mean $T$ was finite. So $t_0$ has at least one child with infinitely many descendants. Thus, we may pick $t_1$ as any one of those children. Now, suppose node $t_k$ has infinitely many descendants.  As $t_k$ has a finite number of children, by the same argument as above, $t_k$ has at least one child with infinitely many descendants. Thus we may pick $t_{k+1}$ which has infinitely many descendants. The assertion follows by the Axiom of Dependent Choice. {{qed}}
Suppose $T$ is a labeled tree with root $r$. Let $L_0 = \set r, L_1, L_2, \ldots$ be the levels of $T$. We construct our path by starting with $r$. Next we partition all vertices in levels higher then $L_0$ (that is, all vertices in levels $L_i$ where $i > 0$) into as many finite parts as is the cardinality of $L_1$. In other words if $L_1 = \set {v_1, \ldots, v_n}$ then we partition the vertices (there are infinitely many of them) in $n$ parts: :$P_1, P_2, \ldots, P_n$ This partitioning is done as follows: Any vertex $w$ will be connected to $r$ via some $v_i$ owing to connectedness. We put $w$ in $P_i$.  Now since we have partitioned infinitely many vertices in finitely many parts, one of the parts will contain infinitely many vertices. (This is also called the Infinite Pigeonhole Principle). Let that part be $P_j$. Choose $w_1 = v_j$ as the next vertex in our under-construction path. Next we delete the vertex $r$ and consider the component containing $w_1$. It is also an infinite tree with each level finite and having $w_1$ as its root. By a similar piece of reasoning, we get a vertex $w_2$ which is adjacent to $w_1$ and has infinitely many vertices corresponding to it. We continue building our path by choosing $w_2$ as a part of it. By induction we can continue and get an infinite path $r w_1 w_2 \cdots$. This proves the theorem. {{qed}}
Suppose $\mathbf H$ does '''not''' have a model. By the Main Lemma of Propositional Tableaus, $\mathbf H$ has a tableau confutation $T$. By Tableau Confutation contains Finite Tableau Confutation, $T$ may be assumed to be finite. Hence the set $\mathbf H'$ of all WFFs in $\mathbf H$ used somewhere in $T$ is finite. Now, let $T'$ be the labeled tree which is the same as $T$ but with root $\mathbf H'$ instead of $\mathbf H$. Then $T'$ is a tableau confutation of $\mathbf H'$. By the Tableau Confutation implies Unsatisfiable, $\mathbf H'$ has no models. But this contradicts the assumption that all finite subsets of $\mathbf H$ have models. Hence the result. {{qed}}
If $\mathbf H$ is finite, the result is trivial. So let $\mathbf H = \left\{{\mathbf A_n: n \in \N}\right\}$ be an enumeration of $\mathbf H$. Define $\mathbf H_m = \left\{{\mathbf A_n: n \le m}\right\}$. Let $T_1$ be the propositional tableau consisting of only a root node with hypothesis set $\mathbf H_1$. For each $m \in \N$, construct $T_{m+1}$ from $T_m$ by applying the Tableau Extension Lemma to $\mathbf H_{m+1}$. This provides us with a sequence $\left({T_m}\right)_{m \in \N}$ of finished propositional tableaus. The $T_m$ form an exhausting sequence of sets for the infinite propositional tableau $T = \displaystyle \bigcup_{m \mathop = 1}^\infty T_m$. If $T$ has a finished branch $\Gamma$, then by Finished Branch Lemma: Corollary: :$\Phi \left[{\Gamma}\right]$ is satisfiable and hence $\mathbf H$ is satisfiable (since $\mathbf H \subseteq \Phi \left[{\Gamma}\right]$). Suppose $T$ has no finite finished branches. Let $T'$ be the subtree of $T$ given by: :$t \in T'$ {{iff}} $t$ is on a finished branch $\Gamma_m$ of every $T_m$ such that $t \in T_m$ Suppose the root node $r_T$ of $T$ were not in $T'$. Then for some $T_m$, $r_T$ would not be on a finished branch of $T_m$. But since $T_m$ is finished, every branch of $T_m$ is finished or contradictory. Hence every branch of $T_m$ is contradictory. But then $T_m$ is a tableau confutation of $\mathbf H_m$. By Tableau Confutation implies Unsatisfiable, this contradicts the assumption that $\mathbf H_m$ is satisfiable. Therefore, $r_T$ is in $T'$. Suppose $T'$ were finite. Let $t$ be a leaf of $T'$, which exists by Finite Tree has Leaf Nodes. Suppose $t$ were a leaf of $T$. Then $\Gamma_t$, the branch of $T$ identified by Leaf of Rooted Tree is on One Branch, is finished. For, as any $T_m$ is a subtree of $T$, $\Gamma_t$ is the only branch of $T_m$ such that $t \in \Gamma_t$. The conclusion follows from the definition of $T'$. But then $\Gamma_t$ would be a finite finished branch of $T$, a contradiction. Therefore, $T'$ cannot be finite. Hence, $T'$ is a finitely branching tree. By König's Tree Lemma, $T'$ has an infinite branch $\Gamma$. By definition of $T'$, the branch $\Gamma_m := \Gamma \cap T_m$ of $T_m$ is finished for each $m \in \N$. If $\Gamma$ were contradictory, then $\mathbf A \in \Gamma$ and $\neg\mathbf A \in \Gamma$ for some WFF $\mathbf A$. But then $\mathbf A, \neg \mathbf A \in \Gamma_m$ for some $m \in \N$, contradicting that $\Gamma_m$ is finished. Also, if $\mathbf A$ occurs on $\Gamma$, then it occurs on some $\Gamma_m$. Since $\Gamma_m$ is finished, it follows that $\mathbf A$ is used on $\Gamma_m$, and hence on $\Gamma$. In conclusion, $\Gamma$ is finished. As established above, it follows that $\mathbf H$ is satisfiable for boolean interpretations. {{qed}}
Let $l \left({\mathbf Q}\right)$ denote the length of a string $\mathbf Q$. By definition, $\mathbf S$ is an initial part of $\mathbf A$ iff $\mathbf A = \mathbf{ST}$ for some non-null string $\mathbf T$. Thus we note that $l \left({\mathbf S}\right) < l \left({\mathbf A}\right)$. The proof proceeds by induction on $l \left({\mathbf A}\right)$. === Basis for the Induction === Let $\mathbf A$ be a WFF such that $l \left({\mathbf A}\right) = 1$. Then for an initial part $\mathbf S$, $l \left({\mathbf S}\right) < 1 = 0$. That is, $\mathbf S$ must be the null string, which is not a WFF. So the result holds for WFFs of length $1$. This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$ with $n \ge n_0$. Assume the result holds for all WFFs of length $k$ or less. This is our induction hypothesis. === Induction Step === This is our induction step: Let $\mathbf A$ be a WFF such that $l \left({\mathbf A}\right) = k+1$. Suppose $\mathbf D$ is an initial part of $\mathbf A$ which happens to be a WFF. That is, $\mathbf A = \mathbf{DT}$ where $\mathbf T$ is non-null. We need to investigate the following cases: :$(1): \quad \mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$. :$(2): \quad \mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ where $\circ$ is one of the binary connectives. :$(3): \quad \mathbf A = p \left({t_1, t_2, \ldots, t_n}\right)$, where $t_1, t_2, \ldots, t_n$ are terms, and $p \in \mathcal{P}_n$. :$(4): \quad \mathbf A = ( Q x: \mathbf B )$, where $\mathbf B$ is a WFF of length $k-5$, $Q$ is a quantifier ($\forall$ or $\exists$) and $x$ is a variable. We deal with these one by one. Cases $(1)$ and $(2)$ are covered by the argument in Initial Part of WFF of PropLog is not WFF. :$(3): \quad$ The atomic WFF $\mathbf A = p \left({u_1, u_2, \ldots, u_n}\right)$: Here we have that $\mathbf D$ would be a string of the form: * $p$ where $p$ is an $n$-ary predicate symbol. This can not be a WFF. * $p ($ which is also not a WFF. * $p (u_1, u_2, \ldots, u_k$ which can not be a WFF. * $p (u_1, u_2, \ldots, u_k,$ which can not be a WFF. :$(4): \quad \mathbf A = ( Q x: \mathbf B )$: $\mathbf D$ can not be $($, $( Q$, $( Q x$ or $( Q x:$ as none of these are WFFs. Thus $\mathbf D$ is a WFF, starting with $( Q x: $; hence $\mathbf D = ( Q x: \mathbf E )$, where $\mathbf E$ is also a WFF. We remove the initial $( Q x: $ and the last closing parenthesis $)$ from $\mathbf A = \mathbf{DT}$ to get $\mathbf B = \mathbf{ET}'$ (where $\mathbf T'$ results from $\mathbf T$ by removing the last closing parenthesis). Recall that $\mathbf B$ is a WFF of length $k-5$. It has $\mathbf E$ as an initial part, which is itself a WFF. This contradicts the induction hypothesis. Therefore no initial part of $\mathbf A = ( Q x: \mathbf{B} )$ can be a WFF. Thus all four cases have been investigated, and we have found that no initial part of any WFF of length $k+1$ can be a WFF. The result follows by the Second Principle of Mathematical Induction. {{qed}}
We have the following cases: # $\mathbf A = \neg \mathbf B$, where $\mathbf B$ is a WFF of length $k$. # $\mathbf A = \left({\mathbf B \circ \mathbf C}\right)$ where $\circ$ is one of the binary connectives. # $\mathbf A = p \left({t_1, t_2, \ldots, t_n}\right)$, where $t_1, t_2, \ldots, t_n$ are terms, and $p \in \mathcal P_n$. # $\mathbf A = ( Q x: \mathbf B )$, where $\mathbf B$ is a WFF of length $k-5$, $Q$ is a quantifier ($\forall$ or $\exists$) and $x$ is a variable. We deal with these in turn. Cases 1 and 2 are taken care of by Language of Propositional Logic has Unique Parsability. Cases 3 and 4 do not start with either a left bracket or a negation sign, so do not have to be investigated.  {{proof wanted|clearly, under the new, correct formal grammar, case 4 ''does'' need investigation}} {{qed}}
=== Existence === First, from the rules of formation of predicate logic, we have that whenever a quantifier is included in a WFF, it appears in the form: :$( Q x: \mathbf B )$ where $\mathbf B$ is itself a WFF. Hence it is clear that $( Q x: \mathbf B )$ is a well-formed part of $\mathbf A$ which begins with $Q$. {{qed|lemma}} === Uniqueness === Now we prove that this well-formed part is unique. Suppose $\mathbf B$ and $\mathbf C$ are both well-formed parts of $\mathbf A$ which begin with the given occurrence of $Q$. Since $\mathbf B$ and $\mathbf C$ both begin with the same $Q$, neither one can be the initial part of the other, by Initial Part of WFF of Predicate Logic is not WFF. Therefore, $\mathbf B$ and $\mathbf C$ are necessarily the same. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = T \subseteq S       | c =  }} {{eqn | ll= \leadsto       | o =        | r = \forall a, b, c \in T: a, b, c \in S       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | o =        | r = a \circ_T \left({b *_T c}\right)       | c =  }} {{eqn | r = a \circ \left({b * c}\right)       | c =  }} {{eqn | r = \left({a \circ b}\right) * \left({a \circ c}\right)       | c = as $\circ$ is distributive over $*$ }} {{eqn | r = \left({a \circ_T b}\right) *_T \left({a \circ_T c}\right)       | c =  }} {{end-eqn}} Similarly for $\left({a *_T b}\right) \circ_T c = \left({a \circ_T c}\right) *_T \left({b \circ_T c}\right)$. {{Qed}} Category:Abstract Algebra Category:Distributive Operations huf47muxnige333w7rktjjaiehw9zfj
{{begin-eqn}} {{eqn | l = p \oplus q       | o = \dashv \vdash       | r = \neg \left ({p \iff q}\right)       | c = Exclusive Or is Negation of Biconditional }} {{eqn | o = \dashv \vdash       | r = \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)       | c = Non-Equivalence as Disjunction of Conjunctions }} {{end-eqn}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccccccc|} \hline p & \oplus & q & (\neg & p & \land & q) & \lor & (p & \land & \neg & q) \\ \hline F & F & F & T & F & F & F & F & F & F & T & F \\ F & T & T & T & F & T & T & T & F & F & F & T  \\ T & T & F & F & T & F & F & T & T & T & T & F  \\ T & F & T & F & T & F & T & F & T & F & F & T  \\ \hline \end{array}$ {{qed}}
{{begin-eqn}} {{eqn | l = \paren {S_1 \cap S_2} \cup \paren {T_1 \cap T_2}       | r = \paren {\paren {S_1 \cap S_2} \cup T_1} \cap \paren {\paren {S_1 \cap S_2} \cup T_2}       | c = Union Distributes over Intersection }} {{eqn | r = \paren {S_1 \cup T_1} \cap \paren {S_2 \cup T_1} \cap \paren {\paren {S_1 \cap S_2} \cup T_2}       | c = Union Distributes over Intersection }} {{eqn | o = \subseteq       | r = S_1 \cup T_1       | c = Intersection is Subset }} {{end-eqn}} {{qed}}
From the definition of symmetric difference, we have: :$R * S = \left({R \setminus S}\right) \cup \left({S \setminus R}\right)$ Thus, expanding: {{begin-eqn}} {{eqn | l = \left({R * S}\right) \cup \left({S * T}\right)       | r = \left({R \setminus S}\right) \cup \left({S \setminus R}\right) \cup \left({S \setminus T}\right) \cup \left({T \setminus S}\right) }} {{eqn | r = \left({\left({R \setminus S}\right) \cup \left({T \setminus S}\right) }\right) \cup \left({\left({S \setminus R}\right) \cup \left({S \setminus T}\right)}\right) }} {{eqn | r = \left({\left({R \cup T}\right) \setminus S}\right) \cup \left({\left({S \setminus R}\right) \cup \left({S \setminus T}\right)}\right)       | c = Set Difference is Right Distributive over Union }} {{eqn | r = \left({\left({R \cup T}\right) \setminus S}\right) \cup \left({S \setminus \left({R \cap T}\right)}\right)       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \left({\left({R \cup S \cup T}\right) \setminus S}\right) \cup \left({S \setminus \left({R \cap T}\right)}\right)       | c = Set Difference with Union is Set Difference }} {{eqn | r = \left({R \cup S \cup T}\right) \setminus \left({R \cap S \cap T}\right)       | c = De Morgan's Laws for Difference with Intersection: Corollary }} {{end-eqn}} {{qed}}
Let $\displaystyle x \in \paren {\bigcup_{\alpha \mathop \in I} S_\alpha} \setminus \paren {\bigcup_{\alpha \mathop \in I} T_\alpha}$. Then by definition of set difference: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcup_{\alpha \mathop \in I} S_\alpha       | c =  }} {{eqn | l = x       | o = \notin       | r = \bigcup_{\alpha \mathop \in I} T_\alpha       | c =  }} {{end-eqn}} By definition of set union, it follows that: {{begin-eqn}} {{eqn | l = \exists \beta \in I: x       | o = \in       | r = S_\beta       | c =  }} {{eqn | l = \neg \exists \beta \in I: x       | o = \in       | r = T_\beta       | c =  }} {{eqn | ll= \leadsto       | l = \forall \beta \in I: x       | o = \notin       | r = T_\beta       | c = De Morgan's Laws (Predicate Logic) }} {{end-eqn}} and so: :$\exists \beta \in I: x \in S_\beta \setminus T_\beta$ Hence: :$\displaystyle x \in \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$ by definition of set union. The result follows by definition of subset. {{qed}}
From Difference of Unions is Subset of Union of Differences, we have: :$\displaystyle \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$ :$\displaystyle \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \mathop \in I} S_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}$ where $\setminus$ denotes set difference. Thus we have: {{begin-eqn}} {{eqn | l = \bigcup_{\alpha \mathop \in I} S_\alpha * \bigcup_{\alpha \mathop \in I} T_\alpha       | r = \paren {\bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} T_\alpha \setminus \bigcup_{\alpha \mathop \in I} S_\alpha}       | c = {{Defof|Symmetric Difference}} }} {{eqn | o = \subseteq       | r = \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha} \cup \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}       | c = Difference of Unions is Subset of Union of Differences }} {{eqn | r = \bigcup_{\alpha \mathop \in I} \paren {\paren {S_\alpha \setminus T_\alpha} \cup \paren {T_\alpha \setminus S_\alpha} }       | c = Union is Associative and Union is Commutative }} {{eqn | r = \bigcup_{\alpha \mathop \in I} \paren {S_\alpha * T_\alpha}       | c = {{Defof|Symmetric Difference}} }} {{end-eqn}} {{qed}} Category:Set Union Category:Symmetric Difference 6cr0258helq86zi4vj1yxlnjy4pft0o
Checking in turn each of the criteria for equivalence: === Reflexive === From the definition, we have that $\forall x \in X: d \left({x, x}\right) = 0$. So $\sim$ is a reflexive relation. {{qed|lemma}} === Symmetric === From the definition, we have that $\forall x, y \in X: d \left({x, y}\right) = d \left({y, x}\right)$. So $\sim$ is a symmetric relation. {{qed|lemma}} === Transitive === From the definition, we have that $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$. So if $d \left({x, y}\right) = 0$ and $d \left({y, z}\right) = 0$ it follows directly that $d \left({x, z}\right) = 0$. So $\sim$ is a transitive relation. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive, and so by definition an equivalence relation. {{qed}} Category:Equivalence Relations Category:Pseudometric Spaces 9ncnsrzc2pse0kumcxulzfxk2ede3wo
From Pseudometric Defines an Equivalence Relation we have that $\sim$ is indeed an equivalence relation. First we verify that $d^*$ is well-defined. Let $z \in \left[\!\left[{x}\right]\!\right]$ and $w \in \left[\!\left[{y}\right]\!\right]$. Then we have: : $\left[\!\left[{z}\right]\!\right] = \left[\!\left[{x}\right]\!\right]$ and: : $\left[\!\left[{w}\right]\!\right] = \left[\!\left[{y}\right]\!\right]$ But: * $d \left({z, w}\right) \le d \left({z, x}\right) + d \left({x, y}\right) + d \left({y, w}\right) \le d \left({x, y}\right)$ * $d \left({x, y}\right) \le d \left({x, z}\right) + d \left({z, w}\right) + d \left({w, y}\right) \le d \left({z, w}\right)$ So $d^*$ is independent of the elements chosen from the equivalence classes. Now we need to show that $d^*$ is a metric. To do that, all we need to do is show that $d^* \left({a, b}\right) > 0$ where $a \ne b$. So let $d^* \left({a, b}\right) = 0$ where $a = \left[\!\left[{x}\right]\!\right], b = \left[\!\left[{y}\right]\!\right]$. Then $d \left({x, y}\right) = 0$ so $y \in \left[\!\left[{x}\right]\!\right]$ and so $a = b$. Hence the result. {{qed}} Category:Pseudometric Spaces Category:Metric Spaces 3q7r4l6rvynvilkm8i10pe511yopuo3
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by: :$-\dfrac {\d x} {\d t} = k x$ for $k \in \R_{>0}$. This is an instance of the Decay Equation, and has the solution: :$x = x_0 e^{-k t}$ {{qed}}
From Differential Equation governing First-Order Reaction, the differential equation governing this reaction is given by: :$-\dfrac {\d x} {\d t} = k x$ for $k \in \R_{>0}$. {{begin-eqn}} {{eqn | l = \int \dfrac {\d x} x       | r = \int - k \rd t       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \ln x       | r = -k t + c       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = e^{- k t + c}       | c =  }} {{eqn | r = e^c e^{- k t}       | c =  }} {{end-eqn}} At time $t = 0$ we have that $x = x_0$. So: :$x_0 = e^c e^0 = e^c$ and hence the result :$x = x_0 e^{-k t}$ {{qed}}
Integrating both sides with respect to $x$: {{begin-eqn}} {{eqn | l = \int \paren {\frac {\d y} {\d x} } \rd x       | r = \int \map f {x, y} \rd x       | c =  }} {{eqn | ll= \leadsto       | l = y + C_1       | r = \int \map f {x, y} \rd x       | c = {{Defof|Indefinite Integral}}: $C_1$ is arbitrary }} {{eqn | ll= \leadsto       | l = y       | r = \int \map f {x, y} \rd x + C       | c = replacing $-C_1$ with $C$ }} {{end-eqn}} The validity of this follows from Picard's Existence Theorem. {{qed}} Category:First Order ODEs mdb7j64ldmykhej6eti5r2kdka1p6u5
From Solution to First Order ODE, the general solution of: :$\dfrac {\d y} {\d x} = \map f {x, y}$ is: :$\displaystyle y = \int \map f {x, \map y x} \rd x + C$ When $x = a$, we have $y = b$. Thus: :$\displaystyle b = \intlimits {\int \map f {x, \map y x} \rd x + C} a {}$ {{MissingLinks|to the notation $[..]_a$}} which gives: :$\displaystyle C = b - \intlimits {\int \map f {x, \map y x} \rd x} a {}$ and so: $\displaystyle y = b + \int \map f {x, \map y x} \rd x - \intlimits {\int \map f {x, \map y x} \rd x} a {}$ whence the result, by the Fundamental Theorem of Calculus. {{qed}} Category:First Order ODEs gt4hxh4a9czxi0upwcmechfb4gvkspj
Integrating both sides of $\size {\map {\phi'} y} \le A$ gives us: {{begin-eqn}}      {{eqn | l = \size {\map {\phi'} y}       | o = \le       | r = A       | c =  }} {{eqn | ll= \leadsto       | l = -A       | o = \le       | r = \map {\phi'} y \le A       | c =  }} {{eqn | ll= \leadsto       | l = \int {-A} \rd y       | o = \le       | r = \map \phi y \le \int A \rd y       | c =  }} {{eqn | ll= \leadsto       | l = -A y       | o = \le       | r = \map \phi y \le y       | c =  }} {{eqn | ll= \leadsto       | l = \size {\map \phi y}       | o = \le       | r = A y       | c =  }} {{end-eqn}} On the interval $\closedint a b$ it follows that $\size {\map \phi y}$ is bounded by the greater of $A a$ and $A b$. Hence the result. {{qed}} Category:Calculus 4u75zt5qhc9g4gat9h520umeozofyws
Dividing both sides by $\map h y$, we get: :$\dfrac 1 {\map h y} \dfrac {\d y} {\d x} = \map g x$ Integrating both sides {{WRT|Integration}} $x$, we get: :$\displaystyle \int \frac 1 {\map h y} \frac {\d y} {\d x} \rd x = \int \map g x \rd x$ which, from Integration by Substitution, reduces to the result. The arbitrary constant $C$ appears during the integration process. {{qed}}
Let: :$\map Q {x, y} = \dfrac {\map M {x, y} } {\map N {x, y} }$ where $M$ and $N$ are homogeneous functions of degree $n$. Let $t \in \R$. Then: {{begin-eqn}} {{eqn | l = \map Q {t x, t y}       | r = \frac {\map M {t x, t y} } {\map N {t x, t y} }       | c =  }} {{eqn | r = \frac {t^n \map M {x, y} } {t^n \map N {x, y} }       | c = as $M$ and $N$ are homogeneous of degree $n$ }} {{eqn | r = t^0 \frac {\map M {x, y} } {\map N {x, y} }       | c =  }} {{eqn | r = t^0 \map Q {x, y}       | c =  }} {{eqn | r = \map Q {x, y}       | c =  }} {{end-eqn}} The result follows from the definition. {{qed}}
From the original equation, we see: :$\dfrac {\d y} {\d x} = \map f {x, y} = -\dfrac {\map M {x, y} } {\map N {x, y} }$ From Quotient of Homogeneous Functions it follows that $\map f {x, y}$ is homogeneous of degree zero. Thus: :$\map f {t x, t y} = t^0 \map f {x, y} = \map f {x, y}$ Set $t = \dfrac 1 x$ in this equation: {{begin-eqn}} {{eqn | l = \map f {x, y}       | r = \map f {\paren {\frac 1 x} x, \paren {\frac 1 x} y}       | c = }} {{eqn | r = \map f {1, \frac y x}       | c = }} {{eqn | r = \map f {1, z}       | c = }} {{end-eqn}} where $z = \dfrac y x$. Then: {{begin-eqn}} {{eqn | l = z       | r = \frac y x       | c = }} {{eqn | ll= \leadsto       | l = y       | r = z x       | c = }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = z + x \frac {\d z} {\d x}       | c = Product Rule }} {{eqn | ll= \leadsto       | l = z + x \frac {\d z} {\d x}       | r = \map f {1, z}       | c = }} {{eqn | ll= \leadsto       | l = \int \frac {\d z} {\map f {1, z} - z}       | r = \int \frac {\d x} x       | c = }} {{end-eqn}} This is seen to be a differential equation with separable variables. On performing the required integrations and simplifying as necessary, the final step is to substitute $\dfrac y x$ back for $z$. {{qed}}