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Theorem eqop 4611
 Description: Express equality to an ordered pair. (Contributed by SF, 6-Jan-2015.)
Assertion
Ref Expression
eqop (A = B, Cz(z A ↔ (t B z = Phi t t C z = ( Phi t ∪ {0c}))))
Distinct variable groups:   z,A,t   z,B,t   t,C,z

Proof of Theorem eqop
StepHypRef Expression
1 dfcleq 2347 . 2 (A = B, Cz(z Az B, C))
2 df-op 4566 . . . . . . 7 B, C = ({z t B z = Phi t} ∪ {z t C z = ( Phi t ∪ {0c})})
32eleq2i 2417 . . . . . 6 (z B, Cz ({z t B z = Phi t} ∪ {z t C z = ( Phi t ∪ {0c})}))
4 elun 3220 . . . . . 6 (z ({z t B z = Phi t} ∪ {z t C z = ( Phi t ∪ {0c})}) ↔ (z {z t B z = Phi t} z {z t C z = ( Phi t ∪ {0c})}))
53, 4bitri 240 . . . . 5 (z B, C ↔ (z {z t B z = Phi t} z {z t C z = ( Phi t ∪ {0c})}))
6 abid 2341 . . . . . 6 (z {z t B z = Phi t} ↔ t B z = Phi t)
7 abid 2341 . . . . . 6 (z {z t C z = ( Phi t ∪ {0c})} ↔ t C z = ( Phi t ∪ {0c}))
86, 7orbi12i 507 . . . . 5 ((z {z t B z = Phi t} z {z t C z = ( Phi t ∪ {0c})}) ↔ (t B z = Phi t t C z = ( Phi t ∪ {0c})))
95, 8bitri 240 . . . 4 (z B, C ↔ (t B z = Phi t t C z = ( Phi t ∪ {0c})))
109bibi2i 304 . . 3 ((z Az B, C) ↔ (z A ↔ (t B z = Phi t t C z = ( Phi t ∪ {0c}))))
1110albii 1566 . 2 (z(z Az B, C) ↔ z(z A ↔ (t B z = Phi t t C z = ( Phi t ∪ {0c}))))
121, 11bitri 240 1 (A = B, Cz(z A ↔ (t B z = Phi t t C z = ( Phi t ∪ {0c}))))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176   ∨ wo 357  ∀wal 1540   = wceq 1642   ∈ wcel 1710  {cab 2339  ∃wrex 2615   ∪ cun 3207  {csn 3737  0cc0c 4374  ⟨cop 4561   Phi cphi 4562 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-un 3214  df-op 4566 This theorem is referenced by:  setconslem3  4733  setconslem7  4737
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