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Theorem eqop2 8014
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 5714 . . 3 𝐵, 𝐶⟩ ∈ (V × V)
4 eleq1 2821 . . 3 (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
53, 4mpbiri 257 . 2 (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6 eqop 8013 . 2 (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
75, 6biadanii 820 1 (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wa 396   = wceq 1541  wcel 2106  Vcvv 3474  cop 4633   × cxp 5673  cfv 6540  1st c1st 7969  2nd c2nd 7970
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2703  ax-sep 5298  ax-nul 5305  ax-pr 5426  ax-un 7721
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-3an 1089  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-mo 2534  df-eu 2563  df-clab 2710  df-cleq 2724  df-clel 2810  df-nfc 2885  df-ne 2941  df-ral 3062  df-rex 3071  df-rab 3433  df-v 3476  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4322  df-if 4528  df-sn 4628  df-pr 4630  df-op 4634  df-uni 4908  df-br 5148  df-opab 5210  df-mpt 5231  df-id 5573  df-xp 5681  df-rel 5682  df-cnv 5683  df-co 5684  df-dm 5685  df-rn 5686  df-iota 6492  df-fun 6542  df-fv 6548  df-1st 7971  df-2nd 7972
This theorem is referenced by:  evlslem4  21628
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