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Proof step 6 from proof ushort-band-01 (previous step)

0 ≤ s2i(sho) + 65536, s2i(sho) < 0 ⊦ (s2i(sho) + 1) % 256 + 255 = (s2i(sho) + 65536) % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
s2i(sho) < 0, ~65536 ≤ s2i(sho) ⊦ (s2i(sho) + 1) % 256 + 255 = (s2i(sho) + 65536) % 256

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;