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Theorem eqop2 7965
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 5673 . . 3 𝐵, 𝐶⟩ ∈ (V × V)
4 eleq1 2826 . . 3 (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
53, 4mpbiri 258 . 2 (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6 eqop 7964 . 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 205  wa 397   = wceq 1542  wcel 2107  Vcvv 3446  cop 4593   × cxp 5632  cfv 6497  1st c1st 7920  2nd c2nd 7921
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-10 2138  ax-11 2155  ax-12 2172  ax-ext 2708  ax-sep 5257  ax-nul 5264  ax-pr 5385  ax-un 7673
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3an 1090  df-tru 1545  df-fal 1555  df-ex 1783  df-nf 1787  df-sb 2069  df-mo 2539  df-eu 2568  df-clab 2715  df-cleq 2729  df-clel 2815  df-nfc 2890  df-ne 2945  df-ral 3066  df-rex 3075  df-rab 3409  df-v 3448  df-dif 3914  df-un 3916  df-in 3918  df-ss 3928  df-nul 4284  df-if 4488  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4867  df-br 5107  df-opab 5169  df-mpt 5190  df-id 5532  df-xp 5640  df-rel 5641  df-cnv 5642  df-co 5643  df-dm 5644  df-rn 5645  df-iota 6449  df-fun 6499  df-fv 6505  df-1st 7922  df-2nd 7923
This theorem is referenced by:  evlslem4  21487
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