Theorem nf4 1868

Description: Variable x is effectively not free in φ iff φ is always true or always false. (Contributed by Mario Carneiro, 24-Sep-2016.)

Assertion

Ref	Expression
nf4	⊢ (Ⅎxφ ↔ (∀xφ ∨ ∀x ¬ φ))

Proof of Theorem nf4

Step	Hyp	Ref	Expression
1		nf2 1866	. 2 ⊢ (Ⅎxφ ↔ (∃xφ → ∀xφ))
2		imor 401	. 2 ⊢ ((∃xφ → ∀xφ) ↔ (¬ ∃xφ ∨ ∀xφ))
3		orcom 376	. . 3 ⊢ ((¬ ∃xφ ∨ ∀xφ) ↔ (∀xφ ∨ ¬ ∃xφ))
4		alnex 1543	. . . 4 ⊢ (∀x ¬ φ ↔ ¬ ∃xφ)
5	4	orbi2i 505	. . 3 ⊢ ((∀xφ ∨ ∀x ¬ φ) ↔ (∀xφ ∨ ¬ ∃xφ))
6	3, 5	bitr4i 243	. 2 ⊢ ((¬ ∃xφ ∨ ∀xφ) ↔ (∀xφ ∨ ∀x ¬ φ))
7	1, 2, 6	3bitri 262	1 ⊢ (Ⅎxφ ↔ (∀xφ ∨ ∀x ¬ φ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357  ∀wal 1540  ∃wex 1541  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-or 359  df-ex 1542  df-nf 1545

This theorem is used by: (None)