# 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 syl5bi
` |-  ( -. 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 ) ) ) )`