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Proof step 5 from proof div-inshort (previous step)

⊦ ¬ s2i(sho0) / s2i(sho1) < ~32768, s2i(sho1) = 0, s2i(sho1) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
s2i(sho0) / s2i(sho1) < ~32768, s2i(sho0) / s2i(sho1) < 0 ⊦ s2i(sho1) = 0, s2i(sho1) = ~1

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1