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Metamath Proof Explorer

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	⊢ ^◡ ( 𝐴 × 𝐵 ) = ( 𝐵 × 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		cnvopab	⊢ ^◡ { ⟨ 𝑦 , 𝑥 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) }
2		ancom	⊢ ( ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴 ) )
3	2	opabbii	⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴 ) }
4	1 3	eqtri	⊢ ^◡ { ⟨ 𝑦 , 𝑥 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴 ) }
5		df-xp	⊢ ( 𝐴 × 𝐵 ) = { ⟨ 𝑦 , 𝑥 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) }
6	5	cnveqi	⊢ ^◡ ( 𝐴 × 𝐵 ) = ^◡ { ⟨ 𝑦 , 𝑥 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) }
7		df-xp	⊢ ( 𝐵 × 𝐴 ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴 ) }
8	4 6 7	3eqtr4i	⊢ ^◡ ( 𝐴 × 𝐵 ) = ( 𝐵 × 𝐴 )