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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 𝑥𝜑
sbbid.2 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
sbbid (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbbid
StepHypRef Expression
1 sbbid.1 . . 3 𝑥𝜑
2 sbbid.2 . . . 4 (𝜑 → (𝜓𝜒))
32biimpd 221 . . 3 (𝜑 → (𝜓𝜒))
41, 3sbimd 2283 . 2 (𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜒))
52biimprd 240 . . 3 (𝜑 → (𝜒𝜓))
61, 5sbimd 2283 . 2 (𝜑 → ([𝑦 / 𝑥]𝜒 → [𝑦 / 𝑥]𝜓))
74, 6impbid 204 1 (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wnf 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
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