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Proof step 6 from proof short-011x-neg (previous step)

s2i(sho) < 0 ∧ 0 ≤ 16, s2i(sho) < 0 ⊦ (s2i(sho) + 1) % 2 ^ 16 + 2 ^ 16 - 1 = s2i(sho) + 65536

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
s2i(sho) < 0 ⊦ (s2i(sho) + 1) % 65536 = 1 + s2i(sho)

Rule argument

Additional information

Used simplifier rules: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;