Theorem sb6x 1659

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)

Hypothesis

Ref	Expression
sb6x.1	⊢ (φ → ∀xφ)

Assertion

Ref	Expression
sb6x	⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 ⊢ (φ → ∀xφ)
2	1	sbh 1656	. 2 ⊢ ([y / x]φ ↔ φ)
3		biidd 161	. . 3 ⊢ (x = y → (φ ↔ φ))
4	1, 3	equsalh 1611	. 2 ⊢ (∀x(x = y → φ) ↔ φ)
5	2, 4	bitr4i 176	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Syntax hints:   → wi 4   ↔ wb 98  ∀wal 1240  [wsb 1642

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-4 1397  ax-i9 1420  ax-ial 1424

This theorem depends on definitions:  df-bi 110  df-sb 1643

This theorem is referenced by: (None)