Theorem nfd 1823

Description: Deduce that 𝑥 is not free in 𝜓 in a context. (Contributed by Wolf Lammen, 16-Sep-2021.)

Hypothesis

Ref	Expression
nfd.1	⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓))

Assertion

Ref	Expression
nfd	⊢ (𝜑 → Ⅎ𝑥𝜓)

Proof of Theorem nfd

Step	Hyp	Ref	Expression
1		nfd.1	. 2 ⊢ (𝜑 → (∃𝑥𝜓 → ∀𝑥𝜓))
2		df-nf 1817	. 2 ⊢ (Ⅎ𝑥𝜓 ↔ (∃𝑥𝜓 → ∀𝑥𝜓))
3	1, 2	sylibr 237	1 ⊢ (𝜑 → Ⅎ𝑥𝜓)

This proof depends on syntax axioms:   → wi 4  ∀wal 1568  ∃wex 1812  Ⅎwnf 1816

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-nf 1817

This theorem is used by:  nftht  1825  nfntht  1826  nfimd  1927  nf5-1  2183  axc16nf  2301  nfald  2363  nfeqf2  2411  bj-nfald  37838