Theorem raleqd 1327

Description: Equality deduction for restricted universal quantifier.

Hypothesis

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

Assertion

Ref	Expression
raleqd	⊢ (A = B → (∀x ∈ A φ ↔ ∀x ∈ B ψ))

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

Proof of Theorem raleqd

Step	Hyp	Ref	Expression
1		raleq 1324	. 2 ⊢ (A = B → (∀x ∈ A φ ↔ ∀x ∈ B φ))
2		raleqd.1	. . 3 ⊢ (A = B → (φ ↔ ψ))
3	2	biraldv 1219	. 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  ∀wral 1201

This theorem is referenced by:  isoeq4 2928  dfom3 3477  aceq1 3552  aceq5lem4 3561  kmlem1 3580  kmlem9 3588  kmlem12 3591  kmlem14 3593  elnp 3886  sh 5116

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-cleq 1097  df-clel 1099  df-ral 1205

metamath.org