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Theorem eqop 8035
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 8032 . . 3 (𝐴 ∈ (𝑉 × 𝑊) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
21eqeq1d 2738 . 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 5456 . 2 (⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶))
62, 5bitrdi 287 1 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1540  wcel 2109  cop 4612   × cxp 5657  cfv 6536  1st c1st 7991  2nd c2nd 7992
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 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2708  ax-sep 5271  ax-nul 5281  ax-pr 5407  ax-un 7734
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2810  df-nfc 2886  df-ne 2934  df-ral 3053  df-rex 3062  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-in 3938  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-uni 4889  df-br 5125  df-opab 5187  df-mpt 5207  df-id 5553  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-rn 5670  df-iota 6489  df-fun 6538  df-fv 6544  df-1st 7993  df-2nd 7994
This theorem is referenced by:  eqop2  8036  op1steq  8037  el2xptp0  8040  lsmhash  19691  txhmeo  23746  ptuncnv  23750  wlkcomp  29616  clwlkcomp  29766  f1od2  32703  gsumwrd2dccatlem  33065  esum2dlem  34128  poimirlem22  37671  rngosn3  37953  dvhb1dimN  41010  f1o2d2  42251
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