Theorem eqop2 8057

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 5725	. . 3 ⊢ ⟨𝐵, 𝐶⟩ ∈ (V × V)
4		eleq1 2829	. . 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 8056	. 2 ⊢ (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))
7	5, 6	biadanii 822	1 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1540   ∈ wcel 2108  Vcvv 3480  ⟨cop 4632   × cxp 5683  ‘cfv 6561  1^st c1st 8012  2^nd c2nd 8013

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 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432  ax-un 7755

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ne 2941  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-uni 4908  df-br 5144  df-opab 5206  df-mpt 5226  df-id 5578  df-xp 5691  df-rel 5692  df-cnv 5693  df-co 5694  df-dm 5695  df-rn 5696  df-iota 6514  df-fun 6563  df-fv 6569  df-1st 8014  df-2nd 8015

This theorem is referenced by:  evlslem4  22100