Metamath Proof Explorer


Theorem abssdv

Description: Deduction of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006) (Proof shortened by SN, 22-Dec-2024)

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

Proof

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