Theorem sb6x 871

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 870	. . 3 ⊢ ([y / x]φ ↔ φ)
3		ax-1 3	. . . 4 ⊢ (φ → (x = y → φ))
4	1, 3	19.21ai 740	. . 3 ⊢ (φ → ∀x(x = y → φ))
5	2, 4	sylbi 174	. 2 ⊢ ([y / x]φ → ∀x(x = y → φ))
6		sb2 859	. 2 ⊢ (∀x(x = y → φ) → [y / x]φ)
7	5, 6	impbi 139	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672   = weq 797  [wsb 852

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-gen 677  ax-9 799

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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