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Metamath Proof Explorer

Theorem ax12indn

Description: Induction step for constructing a substitution instance of ax-c15 without using ax-c15 . Negation case. (Contributed by NM, 21-Jan-2007) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis ax12indn.1 
|- ( -. A. x x = y -> ( x = y -> ( ph -> A. x ( x = y -> ph ) ) ) )
Assertion ax12indn 
|- ( -. A. x x = y -> ( x = y -> ( -. ph -> A. x ( x = y -> -. ph ) ) ) )

Proof

Step Hyp Ref Expression
1 ax12indn.1 
 |-  ( -. A. x x = y -> ( x = y -> ( ph -> A. x ( x = y -> ph ) ) ) )
2 19.8a 
 |-  ( ( x = y /\ -. ph ) -> E. x ( x = y /\ -. ph ) )
3 exanali 
 |-  ( E. x ( x = y /\ -. ph ) <-> -. A. x ( x = y -> ph ) )
4 hbn1 
 |-  ( -. A. x x = y -> A. x -. A. x x = y )
5 hbn1 
 |-  ( -. A. x ( x = y -> ph ) -> A. x -. A. x ( x = y -> ph ) )
6 con3 
 |-  ( ( ph -> A. x ( x = y -> ph ) ) -> ( -. A. x ( x = y -> ph ) -> -. ph ) )
7 1 6 syl6 
 |-  ( -. A. x x = y -> ( x = y -> ( -. A. x ( x = y -> ph ) -> -. ph ) ) )
8 7 com23 
 |-  ( -. A. x x = y -> ( -. A. x ( x = y -> ph ) -> ( x = y -> -. ph ) ) )
9 4 5 8 alrimdh 
 |-  ( -. A. x x = y -> ( -. A. x ( x = y -> ph ) -> A. x ( x = y -> -. ph ) ) )
10 3 9 biimtrid 
 |-  ( -. A. x x = y -> ( E. x ( x = y /\ -. ph ) -> A. x ( x = y -> -. ph ) ) )
11 2 10 syl5 
 |-  ( -. A. x x = y -> ( ( x = y /\ -. ph ) -> A. x ( x = y -> -. ph ) ) )
12 11 expd 
 |-  ( -. A. x x = y -> ( x = y -> ( -. ph -> A. x ( x = y -> -. ph ) ) ) )