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Proof step 164 from proof bytes2int-int2bytes-len-eq (previous step)

b2i(by0) = i + 1, i * 256 + band(b2i(by), 255) < 0, ~129 < i * 256 + band(b2i(by), 255), ~1 ≤ i, 255 * i + band(b2i(by), 255) ≤ 0, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ ¬ i ≤ 0, by = i2b(0), by0 = i2b(0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 165:
b2i(by0) = i + 1, i * 256 + band(b2i(by), 255) < 0, ~129 < i * 256 + band(b2i(by), 255), ~1 ≤ i, 255 * i + band(b2i(by), 255) ≤ 0, i ≤ 0, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by = i2b(0), by0 = i2b(0)

Rule argument

Additional information

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