Theorem cnvxp 5102

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.

Step	Hyp	Ref	Expression
1		cnvopab 5085	. . 3 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
2		ancom 266	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴))
3	2	opabbii 4112	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
4	1, 3	eqtri 2226	. 2 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4682	. . 3 ⊢ (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
6	5	cnveqi 4854	. 2 ⊢ ◡(𝐴 × 𝐵) = ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
7		df-xp 4682	. 2 ⊢ (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
8	4, 6, 7	3eqtr4i 2236	1 ⊢ ◡(𝐴 × 𝐵) = (𝐵 × 𝐴)

Syntax hints:   ∧ wa 104   = wceq 1373   ∈ wcel 2176  {copab 4105   × cxp 4674  ^◡ccnv 4675

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-14 2179  ax-ext 2187  ax-sep 4163  ax-pow 4219  ax-pr 4254

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1484  df-sb 1786  df-eu 2057  df-mo 2058  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-ral 2489  df-rex 2490  df-v 2774  df-un 3170  df-in 3172  df-ss 3179  df-pw 3618  df-sn 3639  df-pr 3640  df-op 3642  df-br 4046  df-opab 4107  df-xp 4682  df-rel 4683  df-cnv 4684

This theorem is referenced by:  xp0  5103  rnxpm  5113  rnxpss  5115  dminxp  5128  imainrect  5129  tposfo  6359  tposf  6360  xpider  6695  xpcomf1o  6922  pw1nct  15977