Theorem eqop2 7959

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 5651	. . 3 ⊢ ⟨𝐵, 𝐶⟩ ∈ (V × V)
4		eleq1 2819	. . 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 7958	. 2 ⊢ (𝐴 ∈ (V × V) → (𝐴 = ⟨𝐵, 𝐶⟩ ↔ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))
7	5, 6	biadanii 821	1 ⊢ (𝐴 = ⟨𝐵, 𝐶⟩ ↔ (𝐴 ∈ (V × V) ∧ ((1st ‘𝐴) = 𝐵 ∧ (2nd ‘𝐴) = 𝐶)))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1541   ∈ wcel 2111  Vcvv 3436  ⟨cop 4577   × cxp 5609  ‘cfv 6476  1^st c1st 7914  2^nd c2nd 7915

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-10 2144  ax-11 2160  ax-12 2180  ax-ext 2703  ax-sep 5229  ax-nul 5239  ax-pr 5365  ax-un 7663

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-mo 2535  df-eu 2564  df-clab 2710  df-cleq 2723  df-clel 2806  df-nfc 2881  df-ne 2929  df-ral 3048  df-rex 3057  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-in 3904  df-ss 3914  df-nul 4279  df-if 4471  df-sn 4572  df-pr 4574  df-op 4578  df-uni 4855  df-br 5087  df-opab 5149  df-mpt 5168  df-id 5506  df-xp 5617  df-rel 5618  df-cnv 5619  df-co 5620  df-dm 5621  df-rn 5622  df-iota 6432  df-fun 6478  df-fv 6484  df-1st 7916  df-2nd 7917

This theorem is referenced by:  evlslem4  22006