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Metamath Proof Explorer

Theorem nf5d

Description: Deduce that x is not free in ps in a context. (Contributed by Mario Carneiro, 24-Sep-2016)

Ref Expression
Hypotheses nf5d.1 ⊢ Ⅎ 𝑥 𝜑
nf5d.2 ⊢ ( 𝜑 → ( 𝜓 → ∀ 𝑥 𝜓 ) )
Assertion nf5d ( 𝜑 → Ⅎ 𝑥 𝜓 )

Proof

Step Hyp Ref Expression
1 nf5d.1 ⊢ Ⅎ 𝑥 𝜑
2 nf5d.2 ⊢ ( 𝜑 → ( 𝜓 → ∀ 𝑥 𝜓 ) )
3 1 2 alrimi ⊢ ( 𝜑 → ∀ 𝑥 ( 𝜓 → ∀ 𝑥 𝜓 ) )
4 nf5-1 ⊢ ( ∀ 𝑥 ( 𝜓 → ∀ 𝑥 𝜓 ) → Ⅎ 𝑥 𝜓 )
5 3 4 syl ⊢ ( 𝜑 → Ⅎ 𝑥 𝜓 )