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Proof step 6 from proof bytes2int-2-le-32767 (previous step)

32768 ≤ band(b2i(by1), 255) + 256 * b2i(by), 0 < band(b2i(by1), 255) + 256 * b2i(by), 32767 < band(b2i(by1), 255) + 256 * b2i(by) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
32768 ≤ band(b2i(by1), 255) + 256 * b2i(by), 0 < band(b2i(by1), 255) + 256 * b2i(by), 0 ≤ band(b2i(by1), 255) + 256 * b2i(by) ⊦

Rule argument

Additional information

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