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Theorem eqop 8024
Description: Two ways to express equality with an ordered pair. (Contributed by NM, 3-Sep-2007.) (Proof shortened by Mario Carneiro, 26-Apr-2015.)
Assertion
Ref Expression
eqop (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))

Proof of Theorem eqop
StepHypRef Expression
1 1st2nd2 8021 . . 3 (𝐴 ∈ (𝑉 × 𝑊) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
21eqeq1d 2765 . 2 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩))
3 fvex 6894 . . 3 (1st𝐴) ∈ V
4 fvex 6894 . . 3 (2nd𝐴) ∈ V
53, 4opth 5458 . 2 (⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶))
62, 5bitrdi 290 1 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 400   = wceq 1570  wcel 2143  cop 4595   × cxp 5659  cfv 6536  1st c1st 7980  2nd c2nd 7981
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-10 2176  ax-11 2192  ax-12 2213  ax-ext 2735  ax-sep 5257  ax-nul 5269  ax-pr 5404  ax-un 7732
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-nf 1814  df-sb 2097  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2838  df-nfc 2912  df-ne 2959  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4873  df-br 5110  df-opab 5174  df-mpt 5193  df-id 5556  df-xp 5667  df-rel 5668  df-cnv 5669  df-co 5670  df-dm 5671  df-rn 5672  df-iota 6492  df-fun 6538  df-fv 6544  df-1st 7982  df-2nd 7983
This theorem is referenced by:  eqop2  8025  op1steq  8026  el2xptp0  8029  mpof1o2d  8117  lsmhash  19770  txhmeo  23960  ptuncnv  23964  wlkcomp  29980  clwlkcomp  30128  f1od2  33064  gsumwrd2dccatlem  33397  esum2dlem  34482  poimirlem22  38293  rngosn3  38575  dvhb1dimN  41760
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