Theorem nfbiOLD 1835

Description: If x is not free in φ and ψ, it is not free in (φ ↔ ψ). (Contributed by Mario Carneiro, 11-Aug-2016.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypotheses

Ref	Expression
nf.1	⊢ Ⅎxφ
nf.2	⊢ Ⅎxψ

Assertion

Ref	Expression
nfbiOLD	⊢ Ⅎx(φ ↔ ψ)

Proof of Theorem nfbiOLD

Step	Hyp	Ref	Expression
1		dfbi2 609	. 2 ⊢ ((φ ↔ ψ) ↔ ((φ → ψ) ∧ (ψ → φ)))
2		nf.1	. . . 4 ⊢ Ⅎxφ
3		nf.2	. . . 4 ⊢ Ⅎxψ
4	2, 3	nfim 1813	. . 3 ⊢ Ⅎx(φ → ψ)
5	3, 2	nfim 1813	. . 3 ⊢ Ⅎx(ψ → φ)
6	4, 5	nfan 1824	. 2 ⊢ Ⅎx((φ → ψ) ∧ (ψ → φ))
7	1, 6	nfxfr 1570	1 ⊢ Ⅎx(φ ↔ ψ)

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-tru 1319  df-ex 1542  df-nf 1545

This theorem is used by: (None)