Theorem eqop2 8012

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 5707	. . 3 ⊢ ⟨𝐵, 𝐶⟩ ∈ (V × V)
4		eleq1 2813	. . 3 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
5	3, 4	mpbiri 258	. 2 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6		eqop 8011	. 2 ⊢ (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))
7	5, 6	biadanii 819	1 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))

Syntax hints:   ↔ wb 205   ∧ wa 395   = wceq 1533   ∈ wcel 2098  Vcvv 3466  ⟨cop 4627   × cxp 5665  ‘cfv 6534  1^st c1st 7967  2^nd c2nd 7968

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2163  ax-ext 2695  ax-sep 5290  ax-nul 5297  ax-pr 5418  ax-un 7719

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-mo 2526  df-eu 2555  df-clab 2702  df-cleq 2716  df-clel 2802  df-nfc 2877  df-ne 2933  df-ral 3054  df-rex 3063  df-rab 3425  df-v 3468  df-dif 3944  df-un 3946  df-in 3948  df-ss 3958  df-nul 4316  df-if 4522  df-sn 4622  df-pr 4624  df-op 4628  df-uni 4901  df-br 5140  df-opab 5202  df-mpt 5223  df-id 5565  df-xp 5673  df-rel 5674  df-cnv 5675  df-co 5676  df-dm 5677  df-rn 5678  df-iota 6486  df-fun 6536  df-fv 6542  df-1st 7969  df-2nd 7970

This theorem is referenced by:  evlslem4  21949