Theorem ceqsrexv 1413

Description: Elimination of a restricted existential quantifier, using implicit substitution.

Hypothesis

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

Assertion

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

Distinct variable group(s):   x,A   x,B   ψ,x

Proof of Theorem ceqsrexv

Step	Hyp	Ref	Expression
1		eleq1 1149	. . . . 5 ⊢ (x = A → (x ∈ B ↔ A ∈ B))
2		ceqsrexv.1	. . . . 5 ⊢ (x = A → (φ ↔ ψ))
3	1, 2	anbi12d 476	. . . 4 ⊢ (x = A → ((x ∈ B ∧ φ) ↔ (A ∈ B ∧ ψ)))
4	3	ceqsexgv 1412	. . 3 ⊢ (A ∈ B → (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ (A ∈ B ∧ ψ)))
5		ibar 487	. . 3 ⊢ (A ∈ B → (ψ ↔ (A ∈ B ∧ ψ)))
6	4, 5	bitr4d 409	. 2 ⊢ (A ∈ B → (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ ψ))
7		df-rex 1206	. . 3 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ ∃x(x ∈ B ∧ (x = A ∧ φ)))
8		an12 370	. . . 4 ⊢ ((x = A ∧ (x ∈ B ∧ φ)) ↔ (x ∈ B ∧ (x = A ∧ φ)))
9	8	biex 733	. . 3 ⊢ (∃x(x = A ∧ (x ∈ B ∧ φ)) ↔ ∃x(x ∈ B ∧ (x = A ∧ φ)))
10	7, 9	bitr4 154	. 2 ⊢ (∃x ∈ B (x = A ∧ φ) ↔ ∃x(x = A ∧ (x ∈ B ∧ φ)))
11	6, 10	syl5bb 410	1 ⊢ (A ∈ B → (∃x ∈ B (x = A ∧ φ) ↔ ψ))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196  ∃wex 678   = wceq 1091   ∈ wcel 1092  ∃wrex 1202

This theorem is referenced by:  reuxfr2 1579  f1oiso 2942

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-rex 1206  df-v 1349

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