Theorem sb6f 2527

Description: Equivalence for substitution when 𝑦 is not free in 𝜑. The implication "to the left" is sb2 2509 and does not require the nonfreeness hypothesis. Theorem sb6 2122 replaces the nonfreeness hypothesis with a disjoint variable condition on 𝑥, 𝑦 and requires fewer axioms. Usage of this theorem is discouraged because it depends on ax-13 2402. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.) (New usage is discouraged.)

Hypothesis

Ref	Expression
sb6f.1	⊢ Ⅎ𝑦𝜑

Assertion

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

Proof of Theorem sb6f

Step	Hyp	Ref	Expression
1		sb6f.1	. . . . 5 ⊢ Ⅎ𝑦𝜑
2	1	nf5ri 2232	. . . 4 ⊢ (𝜑 → ∀𝑦𝜑)
3	2	sbimi 2111	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]∀𝑦𝜑)
4		sb4a 2510	. . 3 ⊢ ([𝑦 / 𝑥]∀𝑦𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
5	3, 4	syl 18	. 2 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
6		sb2 2509	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑)
7	5, 6	impbii 212	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  Ⅎwnf 1816  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2178  ax-12 2213  ax-13 2402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by:  sb5f  2528  bj-sbievv  37740