Theorem sbbid 2284

Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.) Remove dependency on ax-10 2192 and ax-13 2389. (Revised by Wolf Lammen, 24-Nov-2022.)

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	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	2	biimpd 221	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
4	1, 3	sbimd 2283	. 2 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜒))
5	2	biimprd 240	. . 3 ⊢ (𝜑 → (𝜒 → 𝜓))
6	1, 5	sbimd 2283	. 2 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜒 → [𝑦 / 𝑥]𝜓))
7	4, 6	impbid 204	1 ⊢ (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 198  Ⅎwnf 1882  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-12 2220

This theorem depends on definitions:  df-bi 199  df-an 387  df-ex 1879  df-nf 1883  df-sb 2068

This theorem is referenced by:  sbcom3  2542  sbco3  2549  sbcom2  2578  sbcom2OLD  2579  sbal  2596  wl-equsb3  33881  wl-sbcom2d-lem1  33885