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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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