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Theorem cnvxp 6158
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 6141 . . 3 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)}
2 ancom 465 . . . 4 ((𝑦𝐴𝑥𝐵) ↔ (𝑥𝐵𝑦𝐴))
32opabbii 5183 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
41, 3eqtri 2793 . 2 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
5 df-xp 5671 . . 3 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
65cnveqi 5864 . 2 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
7 df-xp 5671 . 2 (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
84, 6, 73eqtr4i 2803 1 (𝐴 × 𝐵) = (𝐵 × 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wa 400   = wceq 1568  wcel 2150  {copab 5178   × cxp 5663  ccnv 5664
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2152  ax-9 2160  ax-11 2199  ax-ext 2742  ax-sep 5262  ax-pr 5408
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1571  df-fal 1581  df-ex 1808  df-sb 2099  df-clab 2749  df-cleq 2762  df-clel 2845  df-rab 3424  df-v 3464  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 5115  df-opab 5179  df-xp 5671  df-rel 5672  df-cnv 5673
This theorem is referenced by:  xp0OLD  6159  rnxp  6172  rnxpss  6174  dminxp  6182  imainrect  6183  cnvrescnv  6198  fparlem3  8112  fparlem4  8113  tposfo  8252  tposf  8253  xpider  8789  xpcomf1o  9057  fpwwe2lem12  10630  trclublem  15035  pjdm  21840  tposmap  22597  ordtrest2  23344  ustneism  24364  trust  24369  metustsym  24695  metust  24698  gtiso  33016  padct  33033  gsumhashmul  33357  ordtcnvNEW  34280  ordtrest2NEW  34283  mbfmcst  34619  eulerpartlemt  34731  0rrv  34811  msrf  35992  mthmpps  36032  elrn3  36212  vxp  38862  trclubgNEW  44296  xpexb  45114  tposresxp  49610  tposf1o  49611
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