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.

Step	Hyp	Ref	Expression
1		cnvopab 4776	. . 3 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
2		ancom 262	. . . 4 ⊢ ((𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴))
3	2	opabbii 3865	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
4	1, 3	eqtri 2103	. 2 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4397	. . 3 ⊢ (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
6	5	cnveqi 4558	. 2 ⊢ ◡(𝐴 × 𝐵) = ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
7		df-xp 4397	. 2 ⊢ (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
8	4, 6, 7	3eqtr4i 2113	1 ⊢ ◡(𝐴 × 𝐵) = (𝐵 × 𝐴)

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