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Theorem eqop 8029
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 8026 . . 3 (𝐴 ∈ (𝑉 × 𝑊) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
21eqeq1d 2730 . 2 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩))
3 fvex 6904 . . 3 (1st𝐴) ∈ V
4 fvex 6904 . . 3 (2nd𝐴) ∈ V
53, 4opth 5472 . 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 205  wa 395   = wceq 1534  wcel 2099  cop 4630   × cxp 5670  cfv 6542  1st c1st 7985  2nd c2nd 7986
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2699  ax-sep 5293  ax-nul 5300  ax-pr 5423  ax-un 7734
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-nf 1779  df-sb 2061  df-mo 2530  df-eu 2559  df-clab 2706  df-cleq 2720  df-clel 2806  df-nfc 2881  df-ne 2937  df-ral 3058  df-rex 3067  df-rab 3429  df-v 3472  df-dif 3948  df-un 3950  df-in 3952  df-ss 3962  df-nul 4319  df-if 4525  df-sn 4625  df-pr 4627  df-op 4631  df-uni 4904  df-br 5143  df-opab 5205  df-mpt 5226  df-id 5570  df-xp 5678  df-rel 5679  df-cnv 5680  df-co 5681  df-dm 5682  df-rn 5683  df-iota 6494  df-fun 6544  df-fv 6550  df-1st 7987  df-2nd 7988
This theorem is referenced by:  eqop2  8030  op1steq  8031  el2xptp0  8034  lsmhash  19653  txhmeo  23700  ptuncnv  23704  wlkcomp  29438  clwlkcomp  29586  f1od2  32497  esum2dlem  33705  poimirlem22  37109  rngosn3  37391  dvhb1dimN  40453  f1o2d2  41718
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