Theorem ceqsex2v 2897

Description: Elimination of two existential quantifiers, using implicit substitution. (Contributed by Scott Fenton, 7-Jun-2006.)

Hypotheses

Ref	Expression
ceqsex2v.1	⊢ A ∈ V
ceqsex2v.2	⊢ B ∈ V
ceqsex2v.3	⊢ (x = A → (φ ↔ ψ))
ceqsex2v.4	⊢ (y = B → (ψ ↔ χ))

Assertion

Ref	Expression
ceqsex2v	⊢ (∃x∃y(x = A ∧ y = B ∧ φ) ↔ χ)

Distinct variable groups:   x,y,A   x,B,y   ψ,x   χ,y

Allowed substitution hints:   φ(x, y)   ψ(y)   χ(x)

Proof of Theorem ceqsex2v

Step	Hyp	Ref	Expression
1		nfv 1619	. 2 ⊢ Ⅎxψ
2		nfv 1619	. 2 ⊢ Ⅎyχ
3		ceqsex2v.1	. 2 ⊢ A ∈ V
4		ceqsex2v.2	. 2 ⊢ B ∈ V
5		ceqsex2v.3	. 2 ⊢ (x = A → (φ ↔ ψ))
6		ceqsex2v.4	. 2 ⊢ (y = B → (ψ ↔ χ))
7	1, 2, 3, 4, 5, 6	ceqsex2 2896	1 ⊢ (∃x∃y(x = A ∧ y = B ∧ φ) ↔ χ)

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ w3a 934  ∃wex 1541   = wceq 1642   ∈ wcel 1710  Vcvv 2860

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 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-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-v 2862

This theorem is used by:  ceqsex3v  2898  ceqsex4v  2899  opksnelsik  4266  sikexlem  4296  br1stg  4731  elswap  4741  brswap2  4861  brsnsi  5774  oqelins4  5795  dmpprod  5841  lecex  6116  addccan2nclem1  6264  nmembers1lem1  6269