Theorem xpcoid 6237

Description: Composition of two Cartesian squares. (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 6209	. . 3 ⊢ (∅ ∘ ∅) = ∅
2		id 22	. . . . . 6 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
3	2	sqxpeqd 5648	. . . . 5 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4		0xp 5715	. . . . 5 ⊢ (∅ × ∅) = ∅
5	3, 4	eqtrdi 2782	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
6	5, 5	coeq12d 5804	. . 3 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
7	1, 6, 5	3eqtr4a 2792	. 2 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8		xpco 6236	. 2 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	7, 8	pm2.61ine 3011	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

Syntax hints:   = wceq 1541  ∅c0 4283   × cxp 5614   ∘ ccom 5620

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-12 2180  ax-ext 2703  ax-sep 5234  ax-nul 5244  ax-pr 5370

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4284  df-if 4476  df-sn 4577  df-pr 4579  df-op 4583  df-br 5092  df-opab 5154  df-xp 5622  df-rel 5623  df-cnv 5624  df-co 5625

This theorem is referenced by:  utop2nei  24163