Metamath Proof Explorer


Theorem ceqsralbv

Description: Elimination of a restricted universal quantifier, using implicit substitution. (Contributed by Scott Fenton, 7-Dec-2020)

Ref Expression
Hypothesis ceqsrexv.1 ⊢ x = A → φ ↔ ψ
Assertion ceqsralbv ⊢ ∀ x ∈ B x = A → φ ↔ A ∈ B → ψ

Proof

Step Hyp Ref Expression
1 ceqsrexv.1 ⊢ x = A → φ ↔ ψ
2 1 notbid ⊢ x = A → ¬ φ ↔ ¬ ψ
3 2 ceqsrexbv ⊢ ∃ x ∈ B x = A ∧ ¬ φ ↔ A ∈ B ∧ ¬ ψ
4 rexanali ⊢ ∃ x ∈ B x = A ∧ ¬ φ ↔ ¬ ∀ x ∈ B x = A → φ
5 annim ⊢ A ∈ B ∧ ¬ ψ ↔ ¬ A ∈ B → ψ
6 3 4 5 3bitr3i ⊢ ¬ ∀ x ∈ B x = A → φ ↔ ¬ A ∈ B → ψ
7 6 con4bii ⊢ ∀ x ∈ B x = A → φ ↔ A ∈ B → ψ