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Proof step 27 from proof div-inshort (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
0 < s2i(sho0) / s2i(sho1), 32767 < s2i(sho0) / s2i(sho1) ⊦ s2i(sho1) = ~1, s2i(sho1) = 0

Rule argument

Additional information

Used simplifier rules: 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