Theorem xpcoid 6135

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 6108	. . 3 ⊢ (∅ ∘ ∅) = ∅
2		id 22	. . . . . 6 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
3	2	sqxpeqd 5581	. . . . 5 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4		0xp 5643	. . . . 5 ⊢ (∅ × ∅) = ∅
5	3, 4	syl6eq 2872	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
6	5, 5	coeq12d 5729	. . 3 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
7	1, 6, 5	3eqtr4a 2882	. 2 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8		xpco 6134	. 2 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	7, 8	pm2.61ine 3100	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

Syntax hints:   = wceq 1533  ∅c0 4290   × cxp 5547   ∘ ccom 5553

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793  ax-sep 5195  ax-nul 5202  ax-pr 5321

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-mo 2618  df-eu 2650  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ne 3017  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4561  df-pr 4563  df-op 4567  df-br 5059  df-opab 5121  df-xp 5555  df-rel 5556  df-cnv 5557  df-co 5558

This theorem is referenced by:  utop2nei  22853