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Theorem eqop 7725
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 7722 . . 3 (𝐴 ∈ (𝑉 × 𝑊) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
21eqeq1d 2823 . 2 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩))
3 fvex 6677 . . 3 (1st𝐴) ∈ V
4 fvex 6677 . . 3 (2nd𝐴) ∈ V
53, 4opth 5360 . 2 (⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶))
62, 5syl6bb 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 398   = wceq 1533  wcel 2110  cop 4566   × cxp 5547  cfv 6349  1st c1st 7681  2nd c2nd 7682
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793  ax-sep 5195  ax-nul 5202  ax-pow 5258  ax-pr 5321  ax-un 7455
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-mo 2618  df-eu 2650  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3496  df-sbc 3772  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4561  df-pr 4563  df-op 4567  df-uni 4832  df-br 5059  df-opab 5121  df-mpt 5139  df-id 5454  df-xp 5555  df-rel 5556  df-cnv 5557  df-co 5558  df-dm 5559  df-rn 5560  df-iota 6308  df-fun 6351  df-fv 6357  df-1st 7683  df-2nd 7684
This theorem is referenced by:  eqop2  7726  op1steq  7727  el2xptp0  7730  lsmhash  18825  txhmeo  22405  ptuncnv  22409  wlkcomp  27406  clwlkcomp  27554  f1od2  30451  esum2dlem  31346  poimirlem22  34908  rngosn3  35196  dvhb1dimN  38116
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