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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.
StepHypRef Expression
1 cnvopab 4898 . . 3 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)}
2 ancom 264 . . . 4 ((𝑦𝐴𝑥𝐵) ↔ (𝑥𝐵𝑦𝐴))
32opabbii 3955 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
41, 3eqtri 2135 . 2 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
5 df-xp 4505 . . 3 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
65cnveqi 4674 . 2 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
7 df-xp 4505 . 2 (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
84, 6, 73eqtr4i 2145 1 (𝐴 × 𝐵) = (𝐵 × 𝐴)
Colors of variables: wff set class
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
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