Theorem sbbid 2230

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.) Remove dependency on ax-10 2129 and ax-13 2363. (Revised by Wolf Lammen, 24-Nov-2022.) Revise df-sb 2060. (Revised by Steven Nguyen, 11-Jul-2023.)

Hypotheses

Ref	Expression
sbbid.1	⊢ Ⅎ𝑥𝜑
sbbid.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbbid	⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbbid

Step	Hyp	Ref	Expression
1		sbbid.1	. . 3 ⊢ Ⅎ𝑥𝜑
2		sbbid.2	. . 3 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	1, 2	alrimi 2198	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
4		spsbbi 2068	. 2 ⊢ (∀𝑥(𝜓 ↔ 𝜒) → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
5	3, 4	syl 17	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1531  Ⅎwnf 1777  [wsb 2059

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-12 2163

This theorem depends on definitions:  df-bi 206  df-ex 1774  df-nf 1778  df-sb 2060

This theorem is referenced by:  2sbbid  2231  sbcom3  2497  sbco3  2504  wl-sbcom2d-lem1  36927  wl-2spsbbi  36933  wl-clabt  36963