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Theorem sbbid 1768
Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.)
Hypotheses
Ref Expression
sbbid.1 𝑥𝜑
sbbid.2 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
sbbid (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbbid
StepHypRef Expression
1 sbbid.1 . . 3 𝑥𝜑
2 sbbid.2 . . 3 (𝜑 → (𝜓𝜒))
31, 2alrimi 1456 . 2 (𝜑 → ∀𝑥(𝜓𝜒))
4 spsbbi 1766 . 2 (∀𝑥(𝜓𝜒) → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
53, 4syl 14 1 (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 103  wal 1283  wnf 1390  [wsb 1686
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-4 1441  ax-ial 1468
This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1687
This theorem is referenced by:  bezoutlemmain  10520
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