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Theorem eqop 8055
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 8052 . . 3 (𝐴 ∈ (𝑉 × 𝑊) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
21eqeq1d 2737 . 2 (𝐴 ∈ (𝑉 × 𝑊) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ⟨(1st𝐴), (2nd𝐴)⟩ = ⟨𝐵, 𝐶⟩))
3 fvex 6920 . . 3 (1st𝐴) ∈ V
4 fvex 6920 . . 3 (2nd𝐴) ∈ V
53, 4opth 5487 . 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 1537  wcel 2106  cop 4637   × cxp 5687  cfv 6563  1st c1st 8011  2nd c2nd 8012
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 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438  ax-un 7754
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-mo 2538  df-eu 2567  df-clab 2713  df-cleq 2727  df-clel 2814  df-nfc 2890  df-ne 2939  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-opab 5211  df-mpt 5232  df-id 5583  df-xp 5695  df-rel 5696  df-cnv 5697  df-co 5698  df-dm 5699  df-rn 5700  df-iota 6516  df-fun 6565  df-fv 6571  df-1st 8013  df-2nd 8014
This theorem is referenced by:  eqop2  8056  op1steq  8057  el2xptp0  8060  lsmhash  19738  txhmeo  23827  ptuncnv  23831  wlkcomp  29664  clwlkcomp  29812  f1od2  32739  gsumwrd2dccatlem  33052  esum2dlem  34073  poimirlem22  37629  rngosn3  37911  dvhb1dimN  40969  f1o2d2  42253
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