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Proof step 15 from proof bytes2int-size-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
2 ^ (n→i(# bytes1) * 8) ≤ bytes2int(by ' + bytes1), 0 < b2i(by), band(255, b2i(by0)) + 256 * bytes2int(by ' + bytes1) < 256 * 2 ^ (8 * n→i(# bytes1)), ∀ by, bytes0. # bytes0 < #(bytes1 + by0 ') → bytes2int(by ' + bytes0) < 2 ^ (n→i(# bytes0) * 8) ∧ 0 < b2i(by) → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ n→i(n);% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);