Metamath Proof Explorer


Theorem axc5c4c711toc5

Description: Rederivation of sp from axc5c4c711 . Note that ax6 is used for the rederivation. (Contributed by Andrew Salmon, 14-Jul-2011) Revised to use ax6v instead of ax6 , so that this rederivation requires only ax6v and propositional calculus. (Revised by BJ, 14-Sep-2019) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion axc5c4c711toc5 ( ∀ 𝑥 𝜑𝜑 )

Proof

Step Hyp Ref Expression
1 ax6v ¬ ∀ 𝑥 ¬ 𝑥 = 𝑦
2 pm2.21 ( ¬ 𝜑 → ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → ¬ 𝑥 = 𝑦 ) ) )
3 ax-1 ( ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → ¬ 𝑥 = 𝑦 ) ) → ( ∀ 𝑥𝑥 ¬ ∀ 𝑥𝑥 ( ∀ 𝑥 𝜑 → ¬ 𝑥 = 𝑦 ) → ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → ¬ 𝑥 = 𝑦 ) ) ) )
4 axc5c4c711 ( ( ∀ 𝑥𝑥 ¬ ∀ 𝑥𝑥 ( ∀ 𝑥 𝜑 → ¬ 𝑥 = 𝑦 ) → ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → ¬ 𝑥 = 𝑦 ) ) ) → ( ∀ 𝑥 𝜑 → ∀ 𝑥 ¬ 𝑥 = 𝑦 ) )
5 2 3 4 3syl ( ¬ 𝜑 → ( ∀ 𝑥 𝜑 → ∀ 𝑥 ¬ 𝑥 = 𝑦 ) )
6 1 5 mtoi ( ¬ 𝜑 → ¬ ∀ 𝑥 𝜑 )
7 6 con4i ( ∀ 𝑥 𝜑𝜑 )