Theorem cnvxp

Description: The converse of a Cartesian product. Exercise 11 of Suppes p. 67. (Contributed by NM, 14-Aug-1999) (Proof shortened by Andrew Salmon, 27-Aug-2011)

	Ref	Expression
Assertion	cnvxp	$$⊢{\left(A\times B\right)}^{-1}=B\times A$$

Proof

Step	Hyp	Ref	Expression
1		cnvopab	$$⊢{\left\{⟨y,x⟩|\left(y\in A\wedge x\in B\right)\right\}}^{-1}=\left\{⟨x,y⟩|\left(y\in A\wedge x\in B\right)\right\}$$
2		ancom	$$⊢\left(y\in A\wedge x\in B\right)\leftrightarrow \left(x\in B\wedge y\in A\right)$$
3	2	opabbii	$$⊢\left\{⟨x,y⟩|\left(y\in A\wedge x\in B\right)\right\}=\left\{⟨x,y⟩|\left(x\in B\wedge y\in A\right)\right\}$$
4	1 3	eqtri	$$⊢{\left\{⟨y,x⟩|\left(y\in A\wedge x\in B\right)\right\}}^{-1}=\left\{⟨x,y⟩|\left(x\in B\wedge y\in A\right)\right\}$$
5		df-xp	$$⊢A\times B=\left\{⟨y,x⟩|\left(y\in A\wedge x\in B\right)\right\}$$
6	5	cnveqi	$$⊢{\left(A\times B\right)}^{-1}={\left\{⟨y,x⟩|\left(y\in A\wedge x\in B\right)\right\}}^{-1}$$
7		df-xp	$$⊢B\times A=\left\{⟨x,y⟩|\left(x\in B\wedge y\in A\right)\right\}$$
8	4 6 7	3eqtr4i	$$⊢{\left(A\times B\right)}^{-1}=B\times A$$