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 φψxA
Assertion abssdv φx|ψA

Proof

Step Hyp Ref Expression
1 abssdv.1 φψxA
2 1 ss2abdv φx|ψx|xA
3 abid1 A=x|xA
4 2 3 sseqtrrdi φx|ψA