Metamath Proof Explorer


Theorem ssabdv

Description: Deduction of abstraction subclass from implication. (Contributed by SN, 22-Dec-2024)

Ref Expression
Hypothesis ssabdv.1 ⊢ φ → x ∈ A → ψ
Assertion ssabdv ⊢ φ → A ⊆ x | ψ

Proof

Step Hyp Ref Expression
1 ssabdv.1 ⊢ φ → x ∈ A → ψ
2 abid1 ⊢ A = x | x ∈ A
3 1 ss2abdv ⊢ φ → x | x ∈ A ⊆ x | ψ
4 2 3 eqsstrid ⊢ φ → A ⊆ x | ψ