Theorem nf5d 2013

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

Hypotheses

Ref	Expression
nf5d.1	⊢ Ⅎ𝑥𝜑
nf5d.2	⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))

Assertion

Ref	Expression
nf5d	⊢ (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nf5d

Step	Hyp	Ref	Expression
1		nf5d.1	. . 3 ⊢ Ⅎ𝑥𝜑
2		nf5d.2	. . 3 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
3	1, 2	alrimi 1510	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4		nf5-1 2012	. 2 ⊢ (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
5	3, 4	syl 14	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1341  Ⅎwnf 1448

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-4 1498  ax-ial 1522  ax-i5r 1523

This theorem depends on definitions:  df-bi 116  df-nf 1449

This theorem is referenced by:  nfabdw  2327