Theorem nfbid 2005

Description: If in a context 𝑥 is not free in 𝜓 and 𝜒, then it is not free in (𝜓 ↔ 𝜒). (Contributed by Mario Carneiro, 24-Sep-2016.) (Proof shortened by Wolf Lammen, 29-Dec-2017.)

Hypotheses

Ref	Expression
nfbid.1	⊢ (𝜑 → Ⅎ𝑥𝜓)
nfbid.2	⊢ (𝜑 → Ⅎ𝑥𝜒)

Assertion

Ref	Expression
nfbid	⊢ (𝜑 → Ⅎ𝑥(𝜓 ↔ 𝜒))

Proof of Theorem nfbid

Step	Hyp	Ref	Expression
1		dfbi2 468	. 2 ⊢ ((𝜓 ↔ 𝜒) ↔ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
2		nfbid.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfbid.2	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	2, 3	nfimd 1996	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → 𝜒))
5	3, 2	nfimd 1996	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜒 → 𝜓))
6	4, 5	nfand 2000	. 2 ⊢ (𝜑 → Ⅎ𝑥((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
7	1, 6	nfxfrd 1953	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386  Ⅎwnf 1882

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-ex 1879  df-nf 1883

This theorem is referenced by:  nfbi  2006  nfeud2OLD  2676  nfeqd  2977  nfiotad  6093  iota2df  6114  axextnd  9735  axrepndlem1  9736  axrepndlem2  9737  axacndlem4  9754  axacndlem5  9755  axacnd  9756  axextdist  32238  wl-eudf  33893  wl-sb8eut  33898