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Proof step 2 from proof short-overflow (previous step)

0 < s2i(sho) + s2i(sho0), 0 ≤ s2i(i2s(s2i(sho) + s2i(sho0))), 32767 < s2i(sho) + s2i(sho0) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i ∧ 32767 < i % 65536 → s2i(i2s(i)) = i % 65536 - 65536

Rule argument

name: sis-pos-high rotate: F subst: [i] → [s2i(sho) + s2i(sho0)]

Additional information

Used simplifier rules: short-mod-add-02: ⊦ 0 < s2i(sho) + s2i(sho0) → (s2i(sho) + s2i(sho0)) % 65536 = s2i(sho) + s2i(sho0);%short-mod-add-02: ⊦ 0 < s2i(sho) + s2i(sho0) → (s2i(sho) + s2i(sho0)) % 65536 = s2i(sho) + s2i(sho0);%short-add-max: ⊦ ¬ 65536 ≤ s2i(sho) + s2i(sho0);% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i + j = j + i;% ⊦ i < j → ¬ j < i;% ⊦ i + j + i0 = (i + j) + i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; Used terms: s2i(sho) + s2i(sho0)