Metamath Proof Explorer


Theorem eqabcdv

Description: Deduction from a wff to a class abstraction. (Contributed by NM, 9-Jul-1994) (Proof shortened by Wolf Lammen, 16-Nov-2019)

Ref Expression
Hypothesis eqabcdv.1 φψxA
Assertion eqabcdv φx|ψ=A

Proof

Step Hyp Ref Expression
1 eqabcdv.1 φψxA
2 1 bicomd φxAψ
3 2 eqabdv φA=x|ψ
4 3 eqcomd φx|ψ=A