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Proof step 8 from proof ushort-band-01 (previous step)

s2i(sho) + 65536 = i * 256 + j ∧ 0 ≤ j ∧ j < 256 ∧ 0 ≤ i ∧ 0 < 256 ∧ 0 ≤ s2i(sho) + 65536, s2i(sho) < 0, ~65536 ≤ s2i(sho) ⊦ (s2i(sho) + 1) % 256 + 255 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
s2i(sho) + 65536 = i * 256 + j, 0 ≤ i, 0 ≤ j, ~65536 ≤ s2i(sho), s2i(sho) < 0, j < 256 ⊦ (s2i(sho) + 1) % 256 + 255 = j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;