ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  eqop2 GIF version

Theorem eqop2 6083
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 4596 . . 3 𝐵, 𝐶⟩ ∈ (V × V)
4 eleq1 2203 . . 3 (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
53, 4mpbiri 167 . 2 (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6 eqop 6082 . 2 (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
75, 6biadan2 452 1 (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st𝐴) = 𝐵 ∧ (2nd𝐴) = 𝐶)))
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104   = wceq 1332  wcel 1481  Vcvv 2689  cop 3534   × cxp 4544  cfv 5130  1st c1st 6043  2nd c2nd 6044
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-13 1492  ax-14 1493  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122  ax-sep 4053  ax-pow 4105  ax-pr 4138  ax-un 4362
This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1737  df-eu 2003  df-mo 2004  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-v 2691  df-sbc 2913  df-un 3079  df-in 3081  df-ss 3088  df-pw 3516  df-sn 3537  df-pr 3538  df-op 3540  df-uni 3744  df-br 3937  df-opab 3997  df-mpt 3998  df-id 4222  df-xp 4552  df-rel 4553  df-cnv 4554  df-co 4555  df-dm 4556  df-rn 4557  df-iota 5095  df-fun 5132  df-fn 5133  df-f 5134  df-fo 5136  df-fv 5138  df-1st 6045  df-2nd 6046
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator