Theorem cnvxp 5155

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

Assertion

Ref	Expression
cnvxp	⊢ ◡(𝐴 × 𝐵) = (𝐵 × 𝐴)

Proof of Theorem cnvxp

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		cnvopab 5138	. . 3 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
2		ancom 266	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴))
3	2	opabbii 4156	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
4	1, 3	eqtri 2252	. 2 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4731	. . 3 ⊢ (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
6	5	cnveqi 4905	. 2 ⊢ ◡(𝐴 × 𝐵) = ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
7		df-xp 4731	. 2 ⊢ (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
8	4, 6, 7	3eqtr4i 2262	1 ⊢ ◡(𝐴 × 𝐵) = (𝐵 × 𝐴)

Syntax hints:   ∧ wa 104   = wceq 1397   ∈ wcel 2202  {copab 4149   × cxp 4723  ^◡ccnv 4724

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-14 2205  ax-ext 2213  ax-sep 4207  ax-pow 4264  ax-pr 4299

This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1811  df-eu 2082  df-mo 2083  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ral 2515  df-rex 2516  df-v 2804  df-un 3204  df-in 3206  df-ss 3213  df-pw 3654  df-sn 3675  df-pr 3676  df-op 3678  df-br 4089  df-opab 4151  df-xp 4731  df-rel 4732  df-cnv 4733

This theorem is referenced by:  xp0  5156  rnxpm  5166  rnxpss  5168  dminxp  5181  imainrect  5182  tposfo  6436  tposf  6437  xpider  6774  xpcomf1o  7008  pw1nct  16604