Let $\phi: \paren {\powerset S} \times \paren {\powerset T} \to  \powerset {S \cup T}$ be defined as: :$\forall \tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}: \map \phi {A, B} = A \cup B$ In order to show that $\phi$ is a bijection, it needs to be shown that $\phi$ has the following properties: :$\phi$ is a mapping, that is: ::$\phi$ is left-total ::$\phi$ is many-to-one :$\phi$ is a surjection :$\phi$ is an injection. Let $A \subseteq S$ and $B \subseteq T$. Then: {{begin-eqn}} {{eqn | l = A       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{eqn | lo= \land       | l = B       | o = \in       | r = \powerset T       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {A, B}       | o = \in       | r = \paren {\powerset S} \times \paren {\powerset T}       | c = {{Defof|Cartesian Product}} }} {{end-eqn}} Thus: :$\forall A \subseteq S, B \subseteq T: \tuple {A, B} \in \Dom \phi$ and so $\phi$ is left-total. {{qed|lemma}} Let $A_1, A_2 \subseteq S$ and $B_1, B_2 \subseteq T$. Then: {{begin-eqn}} {{eqn | l = \map \phi {A_1, B_1}       | o = \ne       | r = \map \phi {A_2, B_2}       | c =  }} {{eqn | ll= \leadsto       | l = A_1 \cup B_1       | o = \ne       | r = A_2 \cup B_2       | c = Definition of $\phi$ }} {{eqn | ll= \leadsto       | l = A_1 \cup B_1       | o = \nsubseteq       | r = A_2 \cup B_2       | c = {{Defof|Set Equality}} }} {{eqn | lo= \lor       | l = A_2 \cup B_2       | o = \nsubseteq       | r = A_1 \cup B_1       | c =  }} {{end-eqn}} {{WLOG}}, suppose $A_1 \cup B_1 \nsubseteq A_2 \cup B_2$. Then: {{begin-eqn}} {{eqn | ll= \exists x \in A_1 \cup B_1:       | l = x       | o = \notin       | r = A_2 \cup B_2       | c =  }} {{eqn | lll=\leadsto       | ll= x \in A_1 \lor x \in B_1:       | l = x \notin A_2       | o = \land       | r = x \notin B_2       | c =  }} {{eqn | lll=\leadsto       | l = A_1 \ne A_2       | o = \lor       | r = A_1 \ne B_2       | c =  }} {{eqn | lo= \land       | l = B_1 \ne A_2       | o = \lor       | r = B_1 \ne B_2       | c =  }} {{eqn | lll=\leadsto       | l = \tuple {A_1, B_1}       | o = \ne       | r = \tuple {A_2, B_2}       | c =  }} {{end-eqn}} That is: :$\map \phi {A_1, B_1} \ne \map \phi {A_2, B_2} \implies \tuple {A_1, B_1} \ne \tuple {A_2, B_2}$ demonstrating that $\phi$ is many-to-one. Thus $\phi$ has been shown to be a mapping. {{qed|lemma}} Let $A \cup B \in \powerset {S \cup T}$. Then: :$\tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}$ by definition of power set and Cartesian product. That is, $\phi$ is a surjection by definition. {{qed|lemma}} It remains to be shown that $\phi$ is an injection. Let $\tuple {A_1, B_1} \ne \tuple {A_2, B_2}$. Then either $A_1 \ne A_2$ or $B_1 \ne B_2$. That is, at least one of the following holds: {{begin-eqn}} {{eqn | l = A_1       | o = \nsubseteq       | r = A_2 }} {{eqn | l = A_2       | o = \nsubseteq       | r = A_1 }} {{eqn | l = B_1       | o = \nsubseteq       | r = B_2 }} {{eqn | l = B_2       | o = \nsubseteq       | r = B_1 }} {{end-eqn}} {{WLOG}}, suppose $A_1 \nsubseteq A_2$. Then: :$\exists x \in A_1: x \notin A_2$ and so by Set is Subset of Union: :$x \in A_1 \cup B_1$ But we have that $S$ and $T$ are disjoint. That is: :$x \in A_1 \implies x \notin B_2$ Thus: :$\exists x \in A_1 \cup B_1: x \notin A_2 \cup B_2$ and so: :$A_1 \cup B_1 \ne A_2 \cup B_2$ That is: :$\map \phi {A_1, B_1} \ne \map \phi {A_2, B_2}$ demonstrating that $\phi$ is an injection. Hence the result. {{qed}}