Theorem xpcoid 5593

Description: Composition of two square Cartesian products. (Contributed by Thierry Arnoux, 14-Jan-2018.)

Assertion

Ref	Expression
xpcoid	⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

Proof of Theorem xpcoid

Step	Hyp	Ref	Expression
1		co01 5567	. . 3 ⊢ (∅ ∘ ∅) = ∅
2		id 22	. . . . . 6 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
3	2	sqxpeqd 5065	. . . . 5 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4		0xp 5122	. . . . 5 ⊢ (∅ × ∅) = ∅
5	3, 4	syl6eq 2660	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
6	5, 5	coeq12d 5208	. . 3 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
7	1, 6, 5	3eqtr4a 2670	. 2 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8		xpco 5592	. 2 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	7, 8	pm2.61ine 2865	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

Syntax hints:   = wceq 1475  ∅c0 3874   × cxp 5036   ∘ ccom 5042

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pr 4833

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-ral 2901  df-rex 2902  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-br 4584  df-opab 4644  df-xp 5044  df-rel 5045  df-cnv 5046  df-co 5047

This theorem is referenced by:  utop2nei  21864