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Proof step 9 from proof bytes2int-size-06 (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 10:
0 ≤ 256 * bytes2int(by ' + bytes1) + band(255, b2i(by0)) + 2 ^ (8 * (n→i(# bytes1) + 1)), b2i(by) < 0, ∀ by, bytes0. # bytes0 < #(bytes1 + by0 ') → 0 ≤ bytes2int(by ' + bytes0) + 2 ^ (8 * n→i(# bytes0)) ∧ b2i(by) < 0 → b2i(by) = ~1 ⊦ 0 ≤ bytes2int(by ' + bytes1) + 2 ^ (8 * n→i(# bytes1)) ∧ b2i(by) < 0, b2i(by) = ~1

Rule argument

subst: [by, bytes0] → [by, bytes1]

Additional information

Used simplifier rules: ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: by, bytes1