Theorem sb6f 1197

Description: Equivalence for substitution when y is not free in φ.

Hypothesis

Ref	Expression
equs45f.1	⊢ (φ → ∀yφ)

Assertion

Ref	Expression
sb6f	⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Proof of Theorem sb6f

Step	Hyp	Ref	Expression
1		equs45f.1	. . . 4 ⊢ (φ → ∀yφ)
2	1	sbimi 1169	. . 3 ⊢ ([y / x]φ → [y / x]∀yφ)
3		sb4a 1195	. . 3 ⊢ ([y / x]∀yφ → ∀x(x = y → φ))
4	2, 3	syl 10	. 2 ⊢ ([y / x]φ → ∀x(x = y → φ))
5		sb2 1173	. 2 ⊢ (∀x(x = y → φ) → [y / x]φ)
6	4, 5	impbi 157	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Syntax hints:   → wi 3   ↔ wb 146  ∀wal 951   = wceq 953  [wsbc 1166

This theorem is referenced by:  sb5f 1198

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 960  ax-11 964  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 978  df-sb 1168

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