Theorem sb6f 2533

Description: Equivalence for substitution when 𝑦 is not free in 𝜑. The implication "to the left" is sb2 2500 and does not require the non-freeness hypothesis. Theorem sb6 2089 replaces the non-freeness hypothesis with a disjoint variable condition and uses less axioms. Usage of this theorem is discouraged because it depends on ax-13 2386. (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 2191	. . . 4 ⊢ (𝜑 → ∀𝑦𝜑)
3	2	sbimi 2075	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]∀𝑦𝜑)
4		sb4a 2505	. . 3 ⊢ ([𝑦 / 𝑥]∀𝑦𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
5	3, 4	syl 17	. 2 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))
6		sb2 2500	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → [𝑦 / 𝑥]𝜑)
7	5, 6	impbii 211	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1531  Ⅎwnf 1780  [wsb 2065

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-10 2141  ax-12 2173  ax-13 2386

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1777  df-nf 1781  df-sb 2066

This theorem is referenced by:  sb5f  2534  bj-sbievv  34167