Theorem ceqsrexv 2668

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 2306	. . 3 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ ∃x(x ∈ B ∧ (x = A ∧ φ)))
2		an12 495	. . . 4 ⊢ ((x = A ∧ (x ∈ B ∧ φ)) ↔ (x ∈ B ∧ (x = A ∧ φ)))
3	2	exbii 1493	. . 3 ⊢ (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ ∃x(x ∈ B ∧ (x = A ∧ φ)))
4	1, 3	bitr4i 176	. 2 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ ∃x(x = A ∧ (x ∈ B ∧ φ)))
5		eleq1 2097	. . . . 5 ⊢ (x = A → (x ∈ B ↔ A ∈ B))
6		ceqsrexv.1	. . . . 5 ⊢ (x = A → (φ ↔ ψ))
7	5, 6	anbi12d 442	. . . 4 ⊢ (x = A → ((x ∈ B ∧ φ) ↔ (A ∈ B ∧ ψ)))
8	7	ceqsexgv 2667	. . 3 ⊢ (A ∈ B → (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ (A ∈ B ∧ ψ)))
9	8	bianabs 543	. 2 ⊢ (A ∈ B → (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ ψ))
10	4, 9	syl5bb 181	1 ⊢ (A ∈ B → (∃x ∈ B (x = A ∧ φ) ↔ ψ))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1242  ∃wex 1378   ∈ 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:  ceqsrexbv  2669  ceqsrex2v  2670  f1oiso  5408  creur  7652  creui  7653