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Theorem eqop2 8011
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 5678 . . 3 𝐵, 𝐶⟩ ∈ (V × V)
4 eleq1 2816 . . 3 (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
53, 4mpbiri 258 . 2 (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6 eqop 8010 . 2 (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
75, 6biadanii 821 1 (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395   = wceq 1540  wcel 2109  Vcvv 3447  cop 4595   × cxp 5636  cfv 6511  1st c1st 7966  2nd c2nd 7967
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2701  ax-sep 5251  ax-nul 5261  ax-pr 5387  ax-un 7711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2533  df-eu 2562  df-clab 2708  df-cleq 2721  df-clel 2803  df-nfc 2878  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3406  df-v 3449  df-dif 3917  df-un 3919  df-in 3921  df-ss 3931  df-nul 4297  df-if 4489  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4872  df-br 5108  df-opab 5170  df-mpt 5189  df-id 5533  df-xp 5644  df-rel 5645  df-cnv 5646  df-co 5647  df-dm 5648  df-rn 5649  df-iota 6464  df-fun 6513  df-fv 6519  df-1st 7968  df-2nd 7969
This theorem is referenced by:  evlslem4  21983
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