Theorem nfbid 1832

Description: If in a context x is not free in ψ and χ, 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 609	. 2 ⊢ ((ψ ↔ χ) ↔ ((ψ → χ) ∧ (χ → ψ)))
2		nfbid.1	. . . 4 ⊢ (φ → Ⅎxψ)
3		nfbid.2	. . . 4 ⊢ (φ → Ⅎxχ)
4	2, 3	nfimd 1808	. . 3 ⊢ (φ → Ⅎx(ψ → χ))
5	3, 2	nfimd 1808	. . 3 ⊢ (φ → Ⅎx(χ → ψ))
6	4, 5	nfand 1822	. 2 ⊢ (φ → Ⅎx((ψ → χ) ∧ (χ → ψ)))
7	1, 6	nfxfrd 1571	1 ⊢ (φ → Ⅎx(ψ ↔ χ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  Ⅎwnf 1544

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This proof depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-nf 1545

This theorem is used by:  nfbi  1834  nfeud2  2216  nfeqd  2504  nfiotad  4343  iota2df  4366