Theorem nf5dh 2200

Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Mario Carneiro, 24-Sep-2016.) df-nf 1885 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 1924	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
4		nf5-1 2198	. 2 ⊢ (∀𝑥(𝜓 → ∀𝑥𝜓) → Ⅎ𝑥𝜓)
5	3, 4	syl 17	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1656  Ⅎwnf 1884

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1896  ax-4 1910  ax-10 2194

This theorem depends on definitions:  df-bi 199  df-ex 1881  df-nf 1885

This theorem is referenced by:  nf5dv  2201  hbimd  2332  ax12indalem  35021  ax12inda2ALT  35022