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Theorem sbrbif 1852
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypotheses
Ref Expression
sbrbif.1 (𝜒 → ∀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 1851 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
3 sbrbif.1 . . . 4 (𝜒 → ∀𝑥𝜒)
43sbh 1675 . . 3 ([𝑦 / 𝑥]𝜒𝜒)
54bibi2i 220 . 2 ((𝜓 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓𝜒))
62, 5bitri 177 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 102  wal 1257  [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: (None)
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