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Proof step 14 from proof getInt-theorem (previous step)

¬(band(b2i(by1), 255) i<< 8 < 2 ^ 16 ∧ 0 ≤ 16 ∧ 0 ≤ band(b2i(by1), 255) i<< 8), 0 ≤ 16 ∧ 16 < 32 ⊦ bor(b2i(by3) i<< 24, bor(band(b2i(by2), 255) * 2 ^ 16, bor(band(b2i(by1), 255) i<< 8, band(b2i(by0), 255)))) = bor(bytes2int(by3 ' + by2 ' + by1 ') * 256, band(b2i(by0), 255)), bor(b2i(by3) i<< 24, bor(band(b2i(by2), 255) * 2 ^ 16, band(b2i(by1), 255) i<< 8)) = bytes2int(by3 ' + by2 ' + by1 ') * 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ bor(band(b2i(by0), 255), bor(band(b2i(by1), 255) i<< 8, bor(65536 * band(b2i(by2), 255), b2i(by3) i<< 24))) = bor(bytes2int(by3 ' + by2 ' + by1 ') * 256, band(b2i(by0), 255)), bor(band(b2i(by1), 255) i<< 8, bor(65536 * band(b2i(by2), 255), b2i(by3) i<< 24)) = bytes2int(by3 ' + by2 ' + by1 ') * 256, ¬ 65536 ≤ band(b2i(by1), 255) i<< 8 ∧ ¬ band(b2i(by1), 255) i<< 8 < 0

Rule argument

Additional information

Used simplifier rules: ⊦ bor(i, j) = bor(j, i);% ⊦ bor(bor(i, i0), j0) = bor(i, bor(i0, j0));% ⊦ bor(i, j) ≠ bor(i0, j) → i ≠ i0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ band(i, j) = band(j, i);