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Theorem cnvxp 6155
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 6138 . . 3 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)}
2 ancom 465 . . . 4 ((𝑦𝐴𝑥𝐵) ↔ (𝑥𝐵𝑦𝐴))
32opabbii 5182 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
41, 3eqtri 2792 . 2 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
5 df-xp 5668 . . 3 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
65cnveqi 5861 . 2 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
7 df-xp 5668 . 2 (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
84, 6, 73eqtr4i 2802 1 (𝐴 × 𝐵) = (𝐵 × 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wa 400   = wceq 1567  wcel 2149  {copab 5177   × cxp 5660  ccnv 5661
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-11 2198  ax-ext 2741  ax-sep 5261  ax-pr 5405
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5114  df-opab 5178  df-xp 5668  df-rel 5669  df-cnv 5670
This theorem is referenced by:  xp0OLD  6156  rnxp  6169  rnxpss  6171  dminxp  6179  imainrect  6180  cnvrescnv  6195  fparlem3  8109  fparlem4  8110  tposfo  8249  tposf  8250  xpider  8786  xpcomf1o  9054  fpwwe2lem12  10627  trclublem  15032  pjdm  21826  tposmap  22583  ordtrest2  23330  ustneism  24350  trust  24355  metustsym  24681  metust  24684  gtiso  32987  padct  33004  gsumhashmul  33328  ordtcnvNEW  34255  ordtrest2NEW  34258  mbfmcst  34594  eulerpartlemt  34706  0rrv  34786  msrf  35967  mthmpps  36007  elrn3  36187  vxp  38836  trclubgNEW  44270  xpexb  45088  tposresxp  49580  tposf1o  49581
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