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Proof step 13 from proof mod-inshort (previous step)

0 < s2i(sho0) % s2i(sho1), 32767 < s2i(sho0) % s2i(sho1) ⊦ s2i(sho1) ≤ 32767 ∧ 0 < s2i(sho1), s2i(sho1) = 0

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

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