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Theorem sbbidh 1799
Description: Deduction substituting both sides of a biconditional. New proofs should use sbbid 1800 instead. (Contributed by NM, 5-Aug-1993.) (New usage is discouraged.)
Hypotheses
Ref Expression
sbbidh.1 (𝜑 → ∀𝑥𝜑)
sbbidh.2 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
sbbidh (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbbidh
StepHypRef Expression
1 sbbidh.1 . . 3 (𝜑 → ∀𝑥𝜑)
2 sbbidh.2 . . 3 (𝜑 → (𝜓𝜒))
31, 2alrimih 1428 . 2 (𝜑 → ∀𝑥(𝜓𝜒))
4 spsbbi 1798 . 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 104  wal 1312  [wsb 1718
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-4 1470  ax-ial 1497
This theorem depends on definitions:  df-bi 116  df-sb 1719
This theorem is referenced by:  sbcomxyyz  1921
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