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Theorem sbrbis 1851
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 1849 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2 sbrbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi1i 221 . 2 (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
41, 3bitri 177 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
Colors of variables: wff set class
Syntax hints:  wb 102  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444
This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662
This theorem is referenced by:  sbrbif  1852  sbabel  2219
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