Theorem cnvxp 5084

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 5067	. . 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 4096	. . 3 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
4	1, 3	eqtri 2214	. 2 ⊢ ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
5		df-xp 4665	. . 3 ⊢ (𝐴 × 𝐵) = {⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
6	5	cnveqi 4837	. 2 ⊢ ◡(𝐴 × 𝐵) = ◡{⟨𝑦, 𝑥⟩ ∣ (𝑦 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)}
7		df-xp 4665	. 2 ⊢ (𝐵 × 𝐴) = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐴)}
8	4, 6, 7	3eqtr4i 2224	1 ⊢ ◡(𝐴 × 𝐵) = (𝐵 × 𝐴)

Syntax hints:   ∧ wa 104   = wceq 1364   ∈ wcel 2164  {copab 4089   × cxp 4657  ^◡ccnv 4658

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-14 2167  ax-ext 2175  ax-sep 4147  ax-pow 4203  ax-pr 4238

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1472  df-sb 1774  df-eu 2045  df-mo 2046  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-rex 2478  df-v 2762  df-un 3157  df-in 3159  df-ss 3166  df-pw 3603  df-sn 3624  df-pr 3625  df-op 3627  df-br 4030  df-opab 4091  df-xp 4665  df-rel 4666  df-cnv 4667

This theorem is referenced by:  xp0  5085  rnxpm  5095  rnxpss  5097  dminxp  5110  imainrect  5111  tposfo  6324  tposf  6325  xpider  6660  xpcomf1o  6879  pw1nct  15493