Theorem sbbid 2258

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.) Remove dependency on ax-10 2152 and ax-13 2380. (Revised by Wolf Lammen, 24-Nov-2022.) Revise df-sb 2074. (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 2225	. 2 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
4		spsbbi 2084	. 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 207  ∀wal 1545  Ⅎwnf 1790  [wsb 2073

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-12 2189

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1787  df-nf 1791  df-sb 2074

This theorem is referenced by:  2sbbid  2259  sbcom3  2514  sbco3  2521  wl-sbcom2d-lem1  37937  wl-2spsbbi  37943  wl-clabt  37965