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Proof step 162 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), 0 ≤ b2i(by0), ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 = i2b(0), by = i2b(0), ¬ b2i(by0) ≤ 1 ∧ ¬ 256 ≤ abs(band(b2i(by), 255))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 163:
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)) ⊦ by = i2b(0), by0 = i2b(0), ¬ i ≤ 0 ∧ ¬ 256 ≤ abs(band(b2i(by), 255))

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ (* i2) * i + j < i + j0 ↔ (i2 - 1) * i + j < j0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i + j = j + i;% ⊦ i * j = j * i;