Theorem nfbid 1935

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 480	. 2 ⊢ ((𝜓 ↔ 𝜒) ↔ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
2		nfbid.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfbid.2	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	2, 3	nfimd 1927	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → 𝜒))
5	3, 2	nfimd 1927	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜒 → 𝜓))
6	4, 5	nfand 1930	. 2 ⊢ (𝜑 → Ⅎ𝑥((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
7	1, 6	nfxfrd 1887	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ↔ 𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  Ⅎwnf 1816

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817

This theorem is used by:  nfbi  1936  nfeqd  2932  nfiotadw  6486  nfiotad  6488  iota2df  6514  axextnd  10648  axrepndlem1  10649  axrepndlem2  10650  axacndlem4  10667  axacndlem5  10668  axacnd  10669  axsepg2  35733  axsepg3  35734  axsepg3ALT  35735  axsepg5  35737  axextdist  36483  copsex2d  37980  cbveud  38215  wl-eudf  38424  wl-sb8eut  38430  wl-sb8eutv  38431