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Metamath Proof Explorer

Theorem alexeqg

Description: Two ways to express substitution of A for x in ph . This is the analogue for classes of sbalex . (Contributed by NM, 2-Mar-1995) (Revised by BJ, 27-Apr-2019)

Ref Expression
Assertion alexeqg 
|- ( A e. V -> ( A. x ( x = A -> ph ) <-> E. x ( x = A /\ ph ) ) )

Proof

Step Hyp Ref Expression
1 eqeq2 
 |-  ( y = A -> ( x = y <-> x = A ) )
2 1 anbi1d 
 |-  ( y = A -> ( ( x = y /\ ph ) <-> ( x = A /\ ph ) ) )
3 2 exbidv 
 |-  ( y = A -> ( E. x ( x = y /\ ph ) <-> E. x ( x = A /\ ph ) ) )
4 1 imbi1d 
 |-  ( y = A -> ( ( x = y -> ph ) <-> ( x = A -> ph ) ) )
5 4 albidv 
 |-  ( y = A -> ( A. x ( x = y -> ph ) <-> A. x ( x = A -> ph ) ) )
6 sbalex 
 |-  ( E. x ( x = y /\ ph ) <-> A. x ( x = y -> ph ) )
7 3 5 6 vtoclbg 
 |-  ( A e. V -> ( E. x ( x = A /\ ph ) <-> A. x ( x = A -> ph ) ) )
8 7 bicomd 
 |-  ( A e. V -> ( A. x ( x = A -> ph ) <-> E. x ( x = A /\ ph ) ) )