Theorem ceqsrexbv 2974

Description: Elimination of a restricted existential quantifier, using implicit substitution. (Contributed by Mario Carneiro, 14-Mar-2014.)

Hypothesis

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

Assertion

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

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

Allowed substitution hint:   φ(x)

Proof of Theorem ceqsrexbv

Step	Hyp	Ref	Expression
1		r19.42v 2766	. 2 ⊢ (∃x ∈ B (A ∈ B ∧ (x = A ∧ φ)) ↔ (A ∈ B ∧ ∃x ∈ B (x = A ∧ φ)))
2		eleq1 2413	. . . . . . 7 ⊢ (x = A → (x ∈ B ↔ A ∈ B))
3	2	adantr 451	. . . . . 6 ⊢ ((x = A ∧ φ) → (x ∈ B ↔ A ∈ B))
4	3	pm5.32ri 619	. . . . 5 ⊢ ((x ∈ B ∧ (x = A ∧ φ)) ↔ (A ∈ B ∧ (x = A ∧ φ)))
5	4	bicomi 193	. . . 4 ⊢ ((A ∈ B ∧ (x = A ∧ φ)) ↔ (x ∈ B ∧ (x = A ∧ φ)))
6	5	baib 871	. . 3 ⊢ (x ∈ B → ((A ∈ B ∧ (x = A ∧ φ)) ↔ (x = A ∧ φ)))
7	6	rexbiia 2648	. 2 ⊢ (∃x ∈ B (A ∈ B ∧ (x = A ∧ φ)) ↔ ∃x ∈ B (x = A ∧ φ))
8		ceqsrexv.1	. . . 4 ⊢ (x = A → (φ ↔ ψ))
9	8	ceqsrexv 2973	. . 3 ⊢ (A ∈ B → (∃x ∈ B (x = A ∧ φ) ↔ ψ))
10	9	pm5.32i 618	. 2 ⊢ ((A ∈ B ∧ ∃x ∈ B (x = A ∧ φ)) ↔ (A ∈ B ∧ ψ))
11	1, 7, 10	3bitr3i 266	1 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ (A ∈ B ∧ ψ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   = 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: (None)