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Theorem eqop2 8056
Description: Two ways to express equality with an ordered pair. (Contributed by NM, 25-Feb-2014.)
Hypotheses
Ref Expression
eqop2.1 𝐵 ∈ V
eqop2.2 𝐶 ∈ V
Assertion
Ref Expression
eqop2 (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))

Proof of Theorem eqop2
StepHypRef Expression
1 eqop2.1 . . . 4 𝐵 ∈ V
2 eqop2.2 . . . 4 𝐶 ∈ V
31, 2opelvv 5729 . . 3 𝐵, 𝐶⟩ ∈ (V × V)
4 eleq1 2827 . . 3 (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
53, 4mpbiri 258 . 2 (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6 eqop 8055 . 2 (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
75, 6biadanii 822 1 (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395   = wceq 1537  wcel 2106  Vcvv 3478  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:  evlslem4  22118
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