Theorem eqop2 8078

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) ∧ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))

Proof of Theorem eqop2

Step	Hyp	Ref	Expression
1		eqop2.1	. . . 4 ⊢ 𝐵 ∈ V
2		eqop2.2	. . . 4 ⊢ 𝐶 ∈ V
3	1, 2	opelvv 5698	. . 3 ⊢ ⟨𝐵, 𝐶⟩ ∈ (V × V)
4		eleq1 2852	. . 3 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
5	3, 4	mpbiri 261	. 2 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6		eqop 8077	. 2 ⊢ (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))
7	5, 6	biadanii 834	1 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147  Vcvv 3454  ⟨cop 4840   × cxp 5656  ‘cfv 6566  1^st c1st 8033  2^nd c2nd 8034

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-nul 5266  ax-pr 5398  ax-un 7783

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-br 5109  df-opab 5174  df-mpt 5193  df-id 5553  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-rn 5669  df-iota 6520  df-fun 6568  df-fv 6574  df-1st 8035  df-2nd 8036

This theorem is used by:  evlslem4  22440