Theorem sb6x 2531

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6x.1	⊢ Ⅎ𝑥𝜑

Assertion

Ref	Expression
sb6x	⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 ⊢ Ⅎ𝑥𝜑
2	1	sbf 2527	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		biidd 252	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜑))
4	1, 3	equsal 2446	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜑)
5	2, 4	bitr4i 267	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 196  ∀wal 1629  Ⅎwnf 1856  [wsb 2049

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-12 2203  ax-13 2408

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-ex 1853  df-nf 1858  df-sb 2050

This theorem is referenced by: (None)