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

¬ abs(s2i(sho) + s2i(sho0)) < 65536, 0 ≤ s2i(sho) ⊦ (s2i(sho) + s2i(sho0)) % 65536 = s2i(sho) + s2i(sho0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ 0 ≤ abs(i);% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;