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Proof step 26 from proof div-inshort (previous step)

0 ≤ s2i(sho0) ∧ 0 < s2i(sho1) ∨ s2i(sho0) < 0 ∧ s2i(sho1) < 0, abs(s2i(sho0)) / abs(s2i(sho1)) < ~32768, abs(s2i(sho0)) / abs(s2i(sho1)) < 0 ⊦ s2i(sho1) = 0, s2i(sho1) = ~1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: div-zero-nls-01: ⊦ j ≠ 0 → ¬ abs(i) / abs(j) < 0; from specification int-div