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Theorem cnvxp 6188
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.)
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 6169 . . 3 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)}
2 ancom 460 . . . 4 ((𝑦𝐴𝑥𝐵) ↔ (𝑥𝐵𝑦𝐴))
32opabbii 5233 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
41, 3eqtri 2768 . 2 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
5 df-xp 5706 . . 3 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
65cnveqi 5899 . 2 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
7 df-xp 5706 . 2 (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
84, 6, 73eqtr4i 2778 1 (𝐴 × 𝐵) = (𝐵 × 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wa 395   = wceq 1537  wcel 2108  {copab 5228   × cxp 5698  ccnv 5699
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-11 2158  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pr 5447
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-ss 3993  df-nul 4353  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-br 5167  df-opab 5229  df-xp 5706  df-rel 5707  df-cnv 5708
This theorem is referenced by:  xp0  6189  rnxp  6201  rnxpss  6203  dminxp  6211  imainrect  6212  cnvrescnv  6226  fparlem3  8155  fparlem4  8156  tposfo  8294  tposf  8295  xpider  8846  xpcomf1o  9127  fpwwe2lem12  10711  trclublem  15044  pjdm  21750  tposmap  22484  ordtrest2  23233  ustneism  24253  trust  24259  metustsym  24589  metust  24592  gtiso  32712  padct  32733  gsumhashmul  33040  ordtcnvNEW  33866  ordtrest2NEW  33869  mbfmcst  34224  eulerpartlemt  34336  0rrv  34416  msrf  35510  mthmpps  35550  elrn3  35724  trclubgNEW  43580  xpexb  44423
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