Theorem xpcoid 6311

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 6282	. . 3 ⊢ (∅ ∘ ∅) = ∅
2		id 22	. . . . . 6 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
3	2	sqxpeqd 5720	. . . . 5 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4		0xp 5786	. . . . 5 ⊢ (∅ × ∅) = ∅
5	3, 4	eqtrdi 2790	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
6	5, 5	coeq12d 5877	. . 3 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
7	1, 6, 5	3eqtr4a 2800	. 2 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8		xpco 6310	. 2 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	7, 8	pm2.61ine 3022	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

Syntax hints:   = wceq 1536  ∅c0 4338   × cxp 5686   ∘ ccom 5692

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-12 2174  ax-ext 2705  ax-sep 5301  ax-nul 5311  ax-pr 5437

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1539  df-fal 1549  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-ne 2938  df-ral 3059  df-rex 3068  df-rab 3433  df-v 3479  df-dif 3965  df-un 3967  df-ss 3979  df-nul 4339  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-br 5148  df-opab 5210  df-xp 5694  df-rel 5695  df-cnv 5696  df-co 5697

This theorem is referenced by:  utop2nei  24274