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Proof step 19 from proof div-inshort (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 20:
~ ~32768 < ~(s2i(sho0) / s2i(sho1)), 0 ≤ s2i(sho0), s2i(sho1) < 0, s2i(sho0) / s2i(sho1) < 0, s2i(sho0) / s2i(sho1) < ~32768 ⊦ s2i(sho1) = ~1

Rule argument

fma: ~ ~32768 < ~(s2i(sho0) / s2i(sho1))

Additional information

Used simplifier rules: neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: ~ ~32768 < ~(s2i(sho0) / s2i(sho1))