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Theorem xpcoid 6115
 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 6088 . . 3 (∅ ∘ ∅) = ∅
2 id 22 . . . . . 6 (𝐴 = ∅ → 𝐴 = ∅)
32sqxpeqd 5561 . . . . 5 (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × ∅))
4 0xp 5623 . . . . 5 (∅ × ∅) = ∅
53, 4syl6eq 2871 . . . 4 (𝐴 = ∅ → (𝐴 × 𝐴) = ∅)
65, 5coeq12d 5709 . . 3 (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (∅ ∘ ∅))
71, 6, 53eqtr4a 2881 . 2 (𝐴 = ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
8 xpco 6114 . 2 (𝐴 ≠ ∅ → ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴))
97, 8pm2.61ine 3089 1 ((𝐴 × 𝐴) ∘ (𝐴 × 𝐴)) = (𝐴 × 𝐴)
 Colors of variables: wff setvar class Syntax hints:   = wceq 1537  ∅c0 4267   × cxp 5527   ∘ ccom 5533 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2792  ax-sep 5177  ax-nul 5184  ax-pr 5304 This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-mo 2622  df-eu 2653  df-clab 2799  df-cleq 2813  df-clel 2891  df-nfc 2959  df-ne 3007  df-ral 3130  df-rex 3131  df-rab 3134  df-v 3475  df-dif 3915  df-un 3917  df-in 3919  df-ss 3928  df-nul 4268  df-if 4442  df-sn 4542  df-pr 4544  df-op 4548  df-br 5041  df-opab 5103  df-xp 5535  df-rel 5536  df-cnv 5537  df-co 5538 This theorem is referenced by:  utop2nei  22832
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