Theorem nfd 1456

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

Hypotheses

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

Assertion

Ref	Expression
nfd	⊢ (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nfd

Step	Hyp	Ref	Expression
1		nfd.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2	1	nfri 1452	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
3		nfd.2	. . 3 ⊢ (𝜑 → (𝜓 → ∀𝑥𝜓))
4	2, 3	alrimih 1398	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 → ∀𝑥𝜓))
5		df-nf 1390	. 2 ⊢ (Ⅎ𝑥𝜓 ↔ ∀𝑥(𝜓 → ∀𝑥𝜓))
6	4, 5	sylibr 132	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

Syntax hints:   → wi 4  ∀wal 1282  Ⅎwnf 1389

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-4 1440

This theorem depends on definitions:  df-bi 115  df-nf 1390

This theorem is referenced by:  nfdh  1457  nfrimi  1458  nfnt  1586  cbv1h  1673  nfald  1683  a16nf  1787  dvelimALT  1927  dvelimfv  1928  nfsb4t  1931  hbeud  1963