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Proof step 125 from proof bytes2int-int2bytes-len-eq (previous step)

b2i(by0) < 0, 256 * b2i(by0) + band(b2i(by), 255) < ~128, ~256 < 256 * b2i(by0) + band(b2i(by), 255), 256 * b2i(by0) + band(b2i(by), 255) < 0, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ 0 < 256 * (b2i(by0) + 1) + band(b2i(by), 255), i2b(~1) = by0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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