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Proof step 9 from proof short-mod-01 (previous step)

⊦ ¬ 32767 < s2i(sho) % i, i = 0, s2i(i2s(s2i(sho) % i)) = s2i(sho) % i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 < s2i(sho) % i, 32767 < s2i(sho) % i ⊦ i = 0

Rule argument

Additional information

Used simplifier rules: short-max-10: ⊦ 32767 < i → s2i(sho) ≠ i;%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)