Metamath Proof Explorer


Theorem axc5c4c711toc4

Description: Rederivation of axc4 from axc5c4c711 . Note that only propositional calculus is required for the rederivation. (Contributed by Andrew Salmon, 14-Jul-2011) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion axc5c4c711toc4 ( ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) → ( ∀ 𝑥 𝜑 → ∀ 𝑥 𝜓 ) )

Proof

Step Hyp Ref Expression
1 ax-1 ⊢ ( ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) → ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) ) )
2 ax-1 ⊢ ( ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) ) → ( ∀ 𝑥 ∀ 𝑥 ¬ ∀ 𝑥 ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) → ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) ) ) )
3 axc5c4c711 ⊢ ( ( ∀ 𝑥 ∀ 𝑥 ¬ ∀ 𝑥 ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) → ( 𝜑 → ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) ) ) → ( ∀ 𝑥 𝜑 → ∀ 𝑥 𝜓 ) )
4 1 2 3 3syl ⊢ ( ∀ 𝑥 ( ∀ 𝑥 𝜑 → 𝜓 ) → ( ∀ 𝑥 𝜑 → ∀ 𝑥 𝜓 ) )