Theorem nfbid 1550

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 383	. 2 ⊢ ((𝜓 ↔ 𝜒) ↔ ((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
2		nfbid.1	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜓)
3		nfbid.2	. . . 4 ⊢ (𝜑 → Ⅎ𝑥𝜒)
4	2, 3	nfimd 1547	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜓 → 𝜒))
5	3, 2	nfimd 1547	. . 3 ⊢ (𝜑 → Ⅎ𝑥(𝜒 → 𝜓))
6	4, 5	nfand 1530	. 2 ⊢ (𝜑 → Ⅎ𝑥((𝜓 → 𝜒) ∧ (𝜒 → 𝜓)))
7	1, 6	nfxfrd 1434	1 ⊢ (𝜑 → Ⅎ𝑥(𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  Ⅎwnf 1419

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408  ax-4 1470  ax-ial 1497  ax-i5r 1498

This theorem depends on definitions:  df-bi 116  df-nf 1420

This theorem is referenced by:  nfbi  1551  nfeudv  1990  nfeqd  2270  nfiotadxy  5049  iota2df  5070  bdsepnft  12777  strcollnft  12874