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Proof step 38 from proof div-inshort (previous step)

¬(¬ 32767 < abs(s2i(sho0)) ∧ s2i(sho0) ≤ 0 ∧ s2i(sho1) < 0), 0 < s2i(sho0) / s2i(sho1), 32767 < s2i(sho0) / s2i(sho1), s2i(sho0) < 0, s2i(sho1) < 0 ⊦ 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-ls: ⊦ s2i(sho) < ~32767 ↔ s2i(sho) = ~32768;%numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs