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Proof step 24 from proof div-inshort (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-nls seq: ⊦ ¬ i0 < i ∧ 0 ≤ i ∧ 0 < j → ¬ i0 < i / j subst: [j, i, i0] → [~ s2i(sho1), s2i(sho0), 32768] flags: rewrite

Additional information

Used simplifier rules: short-max-06: ⊦ 32767 ≤ i → ¬ i < s2i(sho);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j); from specification int-div%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; 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%numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lels-04: ⊦ i ≤ j → (¬ ¬ j < i ↔ false); from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: ~ s2i(sho1), s2i(sho0), 32768