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Proof step 4 from proof div-inshort (previous step)

⊦ s2i(sho1) = 0, s2i(sho1) = ~1, ¬ s2i(sho0) / s2i(sho1) < ~32768 ∧ ¬ 32767 < s2i(sho0) / s2i(sho1)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
⊦ ¬ s2i(sho0) / s2i(sho1) < ~32768, s2i(sho1) = 0, s2i(sho1) = ~1
Proof step 27:
⊦ ¬ 32767 < s2i(sho0) / s2i(sho1), s2i(sho1) = 0, s2i(sho1) = ~1

Rule argument

pos: right 3

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