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Proof step 17 from proof bytes2int-size-07 (previous step)

0 ≤ 8 ∧ 0 ≤ 8 * (1 + n→i(# bytes1)) - 1, bytes2int(i2b(~1) ' + by ' + bytes1) + 2 ^ (8 * (n→i(# bytes1) + 1) - 1) < 0, 0 ≤ 2 ^ 8 * 2 ^ (8 * (1 + n→i(# bytes1)) - 1) + band(255, b2i(by0)) + 256 * bytes2int(i2b(~1) ' + by ' + bytes1), 0 ≤ b2i(by), 2 ^ (8 * (n→i(# bytes1) + 1) - 1) < 255 * bytes2int(i2b(~1) ' + by ' + bytes1) + 2 ^ 8 * 2 ^ (8 * (1 + n→i(# bytes1)) - 1) + band(255, b2i(by0)), ∀ by, bytes0. # bytes0 < #(bytes1 + by0 ') → 0 ≤ b2i(by) ∧ 0 ≤ bytes2int(i2b(~1) ' + by ' + bytes0) + 2 ^ (8 * (n→i(# bytes0) + 1) - 1) → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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 ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j < (* i1) * j + i ↔ 0 < (i1 - 1) * j + i;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ n→i(n);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i * j = j * i;% ⊦ i + j = j + i;