Theorem xpcoid 6248

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 6220	. . 3 ⊢ (∅ ∘ ∅) = ∅
2		id 22	. . . . . 6 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
3	2	sqxpeqd 5656	. . . . 5 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4		0xp 5723	. . . . 5 ⊢ (∅ × ∅) = ∅
5	3, 4	eqtrdi 2788	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
6	5, 5	coeq12d 5813	. . 3 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
7	1, 6, 5	3eqtr4a 2798	. 2 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8		xpco 6247	. 2 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	7, 8	pm2.61ine 3016	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

Syntax hints:   = wceq 1542  ∅c0 4274   × cxp 5622   ∘ ccom 5628

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5231  ax-pr 5370

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-ral 3053  df-rex 3063  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-br 5087  df-opab 5149  df-xp 5630  df-rel 5631  df-cnv 5632  df-co 5633

This theorem is referenced by:  utop2nei  24225