Theorem ceqsrexbv 2669

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 2461	. 2 ⊢ (∃x ∈ B (A ∈ B ∧ (x = A ∧ φ)) ↔ (A ∈ B ∧ ∃x ∈ B (x = A ∧ φ)))
2		eleq1 2097	. . . . . . 7 ⊢ (x = A → (x ∈ B ↔ A ∈ B))
3	2	adantr 261	. . . . . 6 ⊢ ((x = A ∧ φ) → (x ∈ B ↔ A ∈ B))
4	3	pm5.32ri 428	. . . . 5 ⊢ ((x ∈ B ∧ (x = A ∧ φ)) ↔ (A ∈ B ∧ (x = A ∧ φ)))
5	4	bicomi 123	. . . 4 ⊢ ((A ∈ B ∧ (x = A ∧ φ)) ↔ (x ∈ B ∧ (x = A ∧ φ)))
6	5	baib 827	. . 3 ⊢ (x ∈ B → ((A ∈ B ∧ (x = A ∧ φ)) ↔ (x = A ∧ φ)))
7	6	rexbiia 2333	. 2 ⊢ (∃x ∈ B (A ∈ B ∧ (x = A ∧ φ)) ↔ ∃x ∈ B (x = A ∧ φ))
8		ceqsrexv.1	. . . 4 ⊢ (x = A → (φ ↔ ψ))
9	8	ceqsrexv 2668	. . 3 ⊢ (A ∈ B → (∃x ∈ B (x = A ∧ φ) ↔ ψ))
10	9	pm5.32i 427	. 2 ⊢ ((A ∈ B ∧ ∃x ∈ B (x = A ∧ φ)) ↔ (A ∈ B ∧ ψ))
11	1, 7, 10	3bitr3i 199	1 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ (A ∈ B ∧ ψ))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242   ∈ wcel 1390  ∃wrex 2301

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-rex 2306  df-v 2553

This theorem is referenced by:  frecsuclem3  5929