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Proof step 12 from proof bytes2int-len-one-02 (previous step)

1 < #(i2b(0) ' + by ' + bytes), b2i(by) < 0, 0 ≤ bytes2int((i2b(0) ' + by ' + bytes).butlast) * 256 + band(b2i((i2b(0) ' + by ' + bytes).last), 255), bytes2int((i2b(0) ' + by ' + bytes).butlast) * 256 + band(b2i((i2b(0) ' + by ' + bytes).last), 255) < 128 ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
0 ≤ band(255, b2i(bytes.last)) + 256 * bytes2int((i2b(0) ' + by ') + bytes.butlast), band(255, b2i(bytes.last)) + 256 * bytes2int((i2b(0) ' + by ') + bytes.butlast) < 128, b2i(by) < 0 ⊦ bytes = []

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;% ⊦ i * j = j * i;% ⊦ band(i, j) = band(j, i);% ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;% ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ (by ' + by0 ' + bytes).last = (by0 ' + bytes).last;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;