Theorem ceqsrexv 2973

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by NM, 30-Apr-2004.)

Hypothesis

Ref	Expression
ceqsrexv.1	⊢ (x = A → (φ ↔ ψ))

Assertion

Ref	Expression
ceqsrexv	⊢ (A ∈ B → (∃x ∈ B (x = A ∧ φ) ↔ ψ))

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

Allowed substitution hint:   φ(x)

Proof of Theorem ceqsrexv

Step	Hyp	Ref	Expression
1		df-rex 2621	. . 3 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ ∃x(x ∈ B ∧ (x = A ∧ φ)))
2		an12 772	. . . 4 ⊢ ((x = A ∧ (x ∈ B ∧ φ)) ↔ (x ∈ B ∧ (x = A ∧ φ)))
3	2	exbii 1582	. . 3 ⊢ (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ ∃x(x ∈ B ∧ (x = A ∧ φ)))
4	1, 3	bitr4i 243	. 2 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ ∃x(x = A ∧ (x ∈ B ∧ φ)))
5		eleq1 2413	. . . . 5 ⊢ (x = A → (x ∈ B ↔ A ∈ B))
6		ceqsrexv.1	. . . . 5 ⊢ (x = A → (φ ↔ ψ))
7	5, 6	anbi12d 691	. . . 4 ⊢ (x = A → ((x ∈ B ∧ φ) ↔ (A ∈ B ∧ ψ)))
8	7	ceqsexgv 2972	. . 3 ⊢ (A ∈ B → (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ (A ∈ B ∧ ψ)))
9	8	bianabs 850	. 2 ⊢ (A ∈ B → (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ ψ))
10	4, 9	syl5bb 248	1 ⊢ (A ∈ B → (∃x ∈ B (x = A ∧ φ) ↔ ψ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∃wex 1541   = wceq 1642   ∈ wcel 1710  ∃wrex 2616

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-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rex 2621  df-v 2862

This theorem is used by:  ceqsrexbv  2974  ceqsrex2v  2975  fnasrn  5418  f1oiso  5500