Theorem reueqd 1329

Description: Equality deduction for restricted uniqueness quantifier.

Hypothesis

Ref	Expression
raleqd.1	⊢ (A = B → (φ ↔ ψ))

Assertion

Ref	Expression
reueqd	⊢ (A = B → (∃!x ∈ A φ ↔ ∃!x ∈ B ψ))

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

Proof of Theorem reueqd

Step	Hyp	Ref	Expression
1		reueq 1326	. 2 ⊢ (A = B → (∃!x ∈ A φ ↔ ∃!x ∈ B φ))
2		raleqd.1	. . 3 ⊢ (A = B → (φ ↔ ψ))
3	2	bireudv 1318	. 2 ⊢ (A = B → (∃!x ∈ B φ ↔ ∃!x ∈ B ψ))
4	1, 3	bitrd 406	1 ⊢ (A = B → (∃!x ∈ A φ ↔ ∃!x ∈ B ψ))

Syntax hints:   → wi 2   ↔ wb 127   = wceq 1091  ∃!wreu 1203

This theorem is referenced by:  aceq1 3552

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-9 799  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-eu 1009  df-cleq 1097  df-clel 1099  df-reu 1207

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