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Proof step 1 from proof short-overflow (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < s2i(sho) + s2i(sho0), 0 ≤ s2i(i2s(s2i(sho) + s2i(sho0))), 32767 < s2i(sho) + s2i(sho0) ⊦

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)