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Theorem ax11f 2131
Description: Basis step for constructing a substitution instance of ax-11o 2080 without using ax-11o 2080. We can start with any formula  ph in which  x is not free. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypothesis
Ref Expression
ax11f.1  |-  ( ph  ->  A. x ph )
Assertion
Ref Expression
ax11f  |-  ( -. 
A. x  x  =  y  ->  ( x  =  y  ->  ( ph  ->  A. x ( x  =  y  ->  ph )
) ) )

Proof of Theorem ax11f
StepHypRef Expression
1 ax11f.1 . . 3  |-  ( ph  ->  A. x ph )
2 ax-1 5 . . 3  |-  ( ph  ->  ( x  =  y  ->  ph ) )
31, 2alrimih 1552 . 2  |-  ( ph  ->  A. x ( x  =  y  ->  ph )
)
43a1ii 24 1  |-  ( -. 
A. x  x  =  y  ->  ( x  =  y  ->  ( ph  ->  A. x ( x  =  y  ->  ph )
) ) )
Colors of variables: wff set class
Syntax hints:   -. wn 3    -> wi 4   A.wal 1527
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 8  ax-gen 1533  ax-5 1544
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