# 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 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( 𝑥 = 𝑦 → ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) ) )
Assertion ax12indn ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( 𝑥 = 𝑦 → ( ¬ 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦 → ¬ 𝜑 ) ) ) )

### Proof

Step Hyp Ref Expression
1 ax12indn.1 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( 𝑥 = 𝑦 → ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) ) )
2 19.8a ( ( 𝑥 = 𝑦 ∧ ¬ 𝜑 ) → ∃ 𝑥 ( 𝑥 = 𝑦 ∧ ¬ 𝜑 ) )
3 exanali ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ ¬ 𝜑 ) ↔ ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
4 hbn1 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ∀ 𝑥 ¬ ∀ 𝑥 𝑥 = 𝑦 )
5 hbn1 ( ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ∀ 𝑥 ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
6 con3 ( ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ) → ( ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ¬ 𝜑 ) )
7 1 6 syl6 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( 𝑥 = 𝑦 → ( ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ¬ 𝜑 ) ) )
8 7 com23 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ( 𝑥 = 𝑦 → ¬ 𝜑 ) ) )
9 4 5 8 alrimdh ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ¬ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦 → ¬ 𝜑 ) ) )
10 3 9 syl5bi ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ∃ 𝑥 ( 𝑥 = 𝑦 ∧ ¬ 𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦 → ¬ 𝜑 ) ) )
11 2 10 syl5 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ( 𝑥 = 𝑦 ∧ ¬ 𝜑 ) → ∀ 𝑥 ( 𝑥 = 𝑦 → ¬ 𝜑 ) ) )
12 11 expd ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( 𝑥 = 𝑦 → ( ¬ 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦 → ¬ 𝜑 ) ) ) )