Theorem cnvxp 4915

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 4898	. . 3 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
2		ancom 264	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴))
3	2	opabbii 3955	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
4	1, 3	eqtri 2135	. 2 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4505	. . 3 ⊢ (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
6	5	cnveqi 4674	. 2 ⊢ ◡(𝐴 × 𝐵) = ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
7		df-xp 4505	. 2 ⊢ (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
8	4, 6, 7	3eqtr4i 2145	1 ⊢ ◡(𝐴 × 𝐵) = (𝐵 × 𝐴)

Syntax hints:   ∧ wa 103   = wceq 1314   ∈ wcel 1463  {copab 3948   × cxp 4497  ^◡ccnv 4498

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-14 1475  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097  ax-sep 4006  ax-pow 4058  ax-pr 4091

This theorem depends on definitions:  df-bi 116  df-3an 947  df-tru 1317  df-nf 1420  df-sb 1719  df-eu 1978  df-mo 1979  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2244  df-ral 2395  df-rex 2396  df-v 2659  df-un 3041  df-in 3043  df-ss 3050  df-pw 3478  df-sn 3499  df-pr 3500  df-op 3502  df-br 3896  df-opab 3950  df-xp 4505  df-rel 4506  df-cnv 4507

This theorem is referenced by:  xp0  4916  rnxpm  4926  rnxpss  4928  dminxp  4941  imainrect  4942  tposfo  6122  tposf  6123  xpider  6454  xpcomf1o  6672