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Theorem sbrbif 2410
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypotheses
Ref Expression
sbrbif.1 𝑥𝜒
sbrbif.2 ([𝑦 / 𝑥]𝜑𝜓)
Assertion
Ref Expression
sbrbif ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓𝜒))

Proof of Theorem sbrbif
StepHypRef Expression
1 sbrbif.2 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
21sbrbis 2409 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
3 sbrbif.1 . . . 4 𝑥𝜒
43sbf 2384 . . 3 ([𝑦 / 𝑥]𝜒𝜒)
54bibi2i 327 . 2 ((𝜓 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓𝜒))
62, 5bitri 264 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 196  wnf 1705  [wsb 1882
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1841  ax-6 1890  ax-7 1937  ax-10 2021  ax-12 2049  ax-13 2250
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1702  df-nf 1707  df-sb 1883
This theorem is referenced by: (None)
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