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Proof step 4 from proof short-mod-max (previous step)

65536 ≤ abs(s2i(sho0) + s2i(sho1)) ⊦ s2i(sho0) + s2i(sho1) = ~65536, (s2i(sho0) + s2i(sho1)) % 65536 = s2i(sho0) + s2i(sho1), s2i(sho0) + s2i(sho1) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [s2i(sho0) + s2i(sho1)] rotate: F spec: int-abs inst: name: Abs-pos

Additional information

Used simplifier rules: short-add-min: ⊦ s2i(sho0) + s2i(sho1) ≤ ~65536 ↔ s2i(sho0) = ~32768 ∧ s2i(sho1) = ~32768;%short-add-min-01: ⊦ s2i(sho0) + s2i(sho1) = ~65536 ↔ s2i(sho0) = ~32768 ∧ s2i(sho1) = ~32768;%short-add-max: ⊦ ¬ 65536 ≤ s2i(sho) + s2i(sho0);% ⊦ ¬ i < j → (i ≠ j ↔ j < i);% ⊦ j ≤ j0 ∧ 0 < j → i % j ≠ j0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: s2i(sho0) + s2i(sho1)