Theorem nf5dh 2148

Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) df-nf 1784 changed. (Revised by Wolf Lammen, 11-Oct-2021.)

Hypotheses

Ref	Expression
nf5dh.1	⊢ (𝜑 → ∀𝑥𝜑)
nf5dh.2	⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))

Assertion

Ref	Expression
nf5dh	⊢ (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nf5dh

Step	Hyp	Ref	Expression
1		nf5dh.1	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
2		nf5dh.2	. . 3 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
3	1, 2	alrimih 1824	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4		nf5-1 2146	. 2 ⊢ (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
5	3, 4	syl 17	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1538  Ⅎwnf 1783

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-10 2142

This theorem depends on definitions:  df-bi 207  df-ex 1780  df-nf 1784

This theorem is referenced by:  nf5dv  2149  hbimd  2299  ax12indalem  38968  ax12inda2ALT  38969