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Theorem sbrbis 2393
 Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypothesis
Ref Expression
sbrbis.1 ([𝑦 / 𝑥]𝜑𝜓)
Assertion
Ref Expression
sbrbis ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbrbis
StepHypRef Expression
1 sbbi 2389 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2 sbrbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi1i 327 . 2 (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
41, 3bitri 263 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 195  [wsb 1867 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-12 2034  ax-13 2234 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868 This theorem is referenced by:  sbrbif  2394  sbabel  2779
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