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Theorem cnvxp 4792
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 4776 . . 3 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)}
2 ancom 262 . . . 4 ((𝑦𝐴𝑥𝐵) ↔ (𝑥𝐵𝑦𝐴))
32opabbii 3865 . . 3 {⟨𝑥, 𝑦⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
41, 3eqtri 2103 . 2 {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
5 df-xp 4397 . . 3 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
65cnveqi 4558 . 2 (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦𝐴𝑥𝐵)}
7 df-xp 4397 . 2 (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥𝐵𝑦𝐴)}
84, 6, 73eqtr4i 2113 1 (𝐴 × 𝐵) = (𝐵 × 𝐴)
Colors of variables: wff set class
Syntax hints:  wa 102   = wceq 1285  wcel 1434  {copab 3858   × cxp 4389  ccnv 4390
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-14 1446  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2065  ax-sep 3916  ax-pow 3968  ax-pr 3992
This theorem depends on definitions:  df-bi 115  df-3an 922  df-tru 1288  df-nf 1391  df-sb 1688  df-eu 1946  df-mo 1947  df-clab 2070  df-cleq 2076  df-clel 2079  df-nfc 2212  df-ral 2358  df-rex 2359  df-v 2612  df-un 2986  df-in 2988  df-ss 2995  df-pw 3402  df-sn 3422  df-pr 3423  df-op 3425  df-br 3806  df-opab 3860  df-xp 4397  df-rel 4398  df-cnv 4399
This theorem is referenced by:  xp0  4793  rnxpm  4802  rnxpss  4804  dminxp  4815  imainrect  4816  tposfo  5940  tposf  5941  xpiderm  6264  xpcomf1o  6390
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