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Proof step 130 from proof bytes2int-int2bytes-len-eq (previous step)

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)) ⊦ b2i(by0) < 0, i2b(~1) = by0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 131:
0 ≤ b2i(by0), 256 * b2i(by0) + band(b2i(by), 255) < 0, ~256 < 256 * b2i(by0) + band(b2i(by), 255), 256 * b2i(by0) + band(b2i(by), 255) < ~128, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ i2b(~1) = by0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (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;