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Theorem cnvxp 6178
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 6159 . . 3 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)}
2 ancom 460 . . . 4 ((𝑦𝐴𝑥𝐵) ↔ (𝑥𝐵𝑦𝐴))
32opabbii 5214 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
41, 3eqtri 2762 . 2 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
5 df-xp 5694 . . 3 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
65cnveqi 5887 . 2 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
7 df-xp 5694 . 2 (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
84, 6, 73eqtr4i 2772 1 (𝐴 × 𝐵) = (𝐵 × 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wa 395   = wceq 1536  wcel 2105  {copab 5209   × cxp 5686  ccnv 5687
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-11 2154  ax-ext 2705  ax-sep 5301  ax-nul 5311  ax-pr 5437
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1539  df-fal 1549  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-rab 3433  df-v 3479  df-dif 3965  df-un 3967  df-ss 3979  df-nul 4339  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-br 5148  df-opab 5210  df-xp 5694  df-rel 5695  df-cnv 5696
This theorem is referenced by:  xp0  6179  rnxp  6191  rnxpss  6193  dminxp  6201  imainrect  6202  cnvrescnv  6216  fparlem3  8137  fparlem4  8138  tposfo  8276  tposf  8277  xpider  8826  xpcomf1o  9099  fpwwe2lem12  10679  trclublem  15030  pjdm  21744  tposmap  22478  ordtrest2  23227  ustneism  24247  trust  24253  metustsym  24583  metust  24586  gtiso  32715  padct  32736  gsumhashmul  33046  ordtcnvNEW  33880  ordtrest2NEW  33883  mbfmcst  34240  eulerpartlemt  34352  0rrv  34432  msrf  35526  mthmpps  35566  elrn3  35741  trclubgNEW  43607  xpexb  44449
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