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Theorem xpcoid 6310
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
StepHypRef Expression
1 co01 6281 . . 3 (∅ ∘ ∅) = ∅
2 id 22 . . . . . 6 (𝐴 = ∅ → 𝐴 = ∅)
32sqxpeqd 5717 . . . . 5 (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4 0xp 5784 . . . . 5 (∅ × ∅) = ∅
53, 4eqtrdi 2793 . . . 4 (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
65, 5coeq12d 5875 . . 3 (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
71, 6, 53eqtr4a 2803 . 2 (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8 xpco 6309 . 2 (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
97, 8pm2.61ine 3025 1 ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)
Colors of variables: wff setvar class
Syntax hints:   = wceq 1540  c0 4333   × cxp 5683  ccom 5689
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-12 2177  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ne 2941  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-xp 5691  df-rel 5692  df-cnv 5693  df-co 5694
This theorem is referenced by:  utop2nei  24259
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