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Proof step 30 from proof div-inshort (previous step)

¬(¬ 32767 < s2i(sho0) ∧ 0 ≤ s2i(sho0) ∧ 0 < s2i(sho1)), 0 < s2i(sho0) / s2i(sho1), 32767 < s2i(sho0) / s2i(sho1) ⊦ s2i(sho1) = ~1, s2i(sho1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: short-ngr: ⊦ ¬ 32767 < s2i(sho);%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs