Theorem xpcoid 6295

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 6265	. . 3 ⊢ (∅ ∘ ∅) = ∅
2		id 23	. . . . . 6 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
3	2	sqxpeqd 5695	. . . . 5 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4		0xp 5762	. . . . 5 ⊢ (∅ × ∅) = ∅
5	3, 4	eqtrdi 2816	. . . 4 ⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
6	5, 5	coeq12d 5852	. . 3 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
7	1, 6, 5	3eqtr4a 2826	. 2 ⊢ (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8		xpco 6294	. 2 ⊢ (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
9	7, 8	pm2.61ine 3043	1 ⊢ ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)

This proof depends on syntax axioms:   = wceq 1570  ∅c0 4286   × cxp 5661   ∘ ccom 5667

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672

This theorem is used by:  utop2nei  24458