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Theorem eqop 8012
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 8009 . . 3 (𝐴 ∈ (𝑉 × 𝑊) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
21eqeq1d 2764 . 2 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩))
3 fvex 6880 . . 3 (1st𝐴) ∈ V
4 fvex 6880 . . 3 (2nd𝐴) ∈ V
53, 4opth 5444 . 2 (⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶))
62, 5bitrdi 289 1 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 399   = wceq 1560  wcel 2142  cop 4588   × cxp 5645  cfv 6521  1st c1st 7968  2nd c2nd 7969
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734  ax-sep 5246  ax-nul 5256  ax-pr 5390  ax-un 7718
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1100  df-tru 1563  df-fal 1573  df-ex 1800  df-nf 1804  df-sb 2091  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3077  df-rex 3087  df-rab 3415  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4481  df-sn 4583  df-pr 4585  df-op 4589  df-uni 4866  df-br 5101  df-opab 5163  df-mpt 5182  df-id 5542  df-xp 5653  df-rel 5654  df-cnv 5655  df-co 5656  df-dm 5657  df-rn 5658  df-iota 6477  df-fun 6523  df-fv 6529  df-1st 7970  df-2nd 7971
This theorem is referenced by:  eqop2  8013  op1steq  8014  el2xptp0  8017  mpof1o2d  8105  lsmhash  19745  txhmeo  23863  ptuncnv  23867  wlkcomp  29831  clwlkcomp  29979  f1od2  32921  gsumwrd2dccatlem  33257  esum2dlem  34389  poimirlem22  38141  rngosn3  38423  dvhb1dimN  41610
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