Theorem sbhb 2107

Description: Two ways of expressing "x is (effectively) not free in φ." (Contributed by NM, 29-May-2009.)

Assertion

Ref	Expression
sbhb	⊢ ((φ → ∀xφ) ↔ ∀y(φ → [y / x]φ))

Distinct variable group:   φ,y

Allowed substitution hint:   φ(x)

Proof of Theorem sbhb

Step	Hyp	Ref	Expression
1		nfv 1619	. . . 4 ⊢ Ⅎyφ
2	1	sb8 2092	. . 3 ⊢ (∀xφ ↔ ∀y[y / x]φ)
3	2	imbi2i 303	. 2 ⊢ ((φ → ∀xφ) ↔ (φ → ∀y[y / x]φ))
4		19.21v 1890	. 2 ⊢ (∀y(φ → [y / x]φ) ↔ (φ → ∀y[y / x]φ))
5	3, 4	bitr4i 243	1 ⊢ ((φ → ∀xφ) ↔ ∀y(φ → [y / x]φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176  ∀wal 1540  [wsb 1648

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-7 1734  ax-11 1746  ax-12 1925

This proof depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by:  sbnf2  2108