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Proof step 22 from proof short-2bytes (previous step)

32767 = band(b2i(by0), 255) + 256 * 127 ∨ band(b2i(by0), 255) + 256 * 127 < 32767, 0 < band(b2i(by0), 255) + 256 * b2i(by), 32767 < band(b2i(by0), 255) + 256 * b2i(by) ⊦

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ 0 < i ∧ j < 0 → ((* j) < (* i) * i0 ↔ (j + 1 - i) / i < i0);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ ¬ 127 < b2i(by);