Theorem sbbid 2281

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.) Remove dependency on ax-10 2177 and ax-13 2401. (Revised by Wolf Lammen, 24-Nov-2022.) Revise df-sb 2099. (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 2249	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
4		spsbbi 2109	. 2 ⊢ (∀𝑥(𝜓 ↔ 𝜒) → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
5	3, 4	syl 18	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  Ⅎwnf 1815  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-sb 2099

This theorem is used by:  2sbbid  2282  sbcom3  2535  sbco3  2542  wl-sbcom2d-lem1  38598  wl-2spsbbi  38604  wl-clabt  38626