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Theorem sbrbis 2314
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 2311 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2 sbrbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi1i 340 . 2 (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
41, 3bitri 276 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 207  [wsb 2062
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-10 2138  ax-12 2169
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 844  df-ex 1774  df-nf 1778  df-sb 2063
This theorem is referenced by:  sbrbif  2315  sbabel  3019
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