Theorem sb6x 1190

Description: Equivalence involving substitution for a variable not free.

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	. . . 4 ⊢ (φ → ∀xφ)
2	1	sbf 1188	. . 3 ⊢ ([y / x]φ ↔ φ)
3		ax-1 4	. . . 4 ⊢ (φ → (x = y → φ))
4	1, 3	19.21ai 1000	. . 3 ⊢ (φ → ∀x(x = y → φ))
5	2, 4	sylbi 199	. 2 ⊢ ([y / x]φ → ∀x(x = y → φ))
6		sb2 1179	. 2 ⊢ (∀x(x = y → φ) → [y / x]φ)
7	5, 6	impbi 157	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Syntax hints:   → wi 3   ↔ wb 146  ∀wal 956   = wceq 958  [wsbc 1172

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 965  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174

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