Theorem eqop2 7726

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 5589	. . 3 ⊢ ⟨𝐵, 𝐶⟩ ∈ (V × V)
4		eleq1 2900	. . 3 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ → (𝐴 ∈ (V × V) ↔ ⟨𝐵, 𝐶⟩ ∈ (V × V)))
5	3, 4	mpbiri 260	. 2 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ → 𝐴 ∈ (V × V))
6		eqop 7725	. 2 ⊢ (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))
7	5, 6	biadanii 820	1 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))

Syntax hints:   ↔ wb 208   ∧ wa 398   = wceq 1533   ∈ wcel 2110  Vcvv 3495  ⟨cop 4567   × cxp 5548  ‘cfv 6350  1^st c1st 7681  2^nd c2nd 7682

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2156  ax-12 2172  ax-ext 2793  ax-sep 5196  ax-nul 5203  ax-pow 5259  ax-pr 5322  ax-un 7455

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-mo 2618  df-eu 2650  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3497  df-sbc 3773  df-dif 3939  df-un 3941  df-in 3943  df-ss 3952  df-nul 4292  df-if 4468  df-sn 4562  df-pr 4564  df-op 4568  df-uni 4833  df-br 5060  df-opab 5122  df-mpt 5140  df-id 5455  df-xp 5556  df-rel 5557  df-cnv 5558  df-co 5559  df-dm 5560  df-rn 5561  df-iota 6309  df-fun 6352  df-fv 6358  df-1st 7683  df-2nd 7684

This theorem is referenced by:  evlslem4  20282