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Proof step 17 from proof short-2bytes (previous step)

⊦ ¬ 32767 < band(b2i(by0), 255) + 256 * b2i(by), s2i(i2s(band(b2i(by0), 255) + 256 * b2i(by))) = band(b2i(by0), 255) + 256 * b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
0 < band(b2i(by0), 255) + 256 * b2i(by), 32767 < band(b2i(by0), 255) + 256 * b2i(by) ⊦

Rule argument

Additional information

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