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Metamath Proof Explorer

Theorem eqab2

Description: Implication of a class abstraction. (Contributed by Peter Mazsa, 16-Apr-2019)

Ref Expression
Assertion eqab2 
|- ( A. x ( x e. A <-> ph ) -> A. x e. A ph )

Proof

Step Hyp Ref Expression
1 biimp 
 |-  ( ( x e. A <-> ph ) -> ( x e. A -> ph ) )
2 1 alimi 
 |-  ( A. x ( x e. A <-> ph ) -> A. x ( x e. A -> ph ) )
3 2 ralrid 
 |-  ( A. x ( x e. A <-> ph ) -> A. x e. A ph )