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Proof step 9 from proof bytes2int-neg (previous step)

bytes1 = bytes + by0 ', ∀ bytes0. # bytes0 < #(by + bytes1) → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ bytes1 = [], band(255, b2i(by0)) + 256 * bytes2int(by ' + bytes) < 0 ↔ ¬ 0 ≤ b2i(by)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ bytes0. # bytes0 < #(by + (bytes + by0 ')) → true → bytes0 ≠ [] → (bytes2int(bytes0) < 0 ↔ b2i(bytes0.first) < 0) ⊦ band(255, b2i(by0)) + 256 * bytes2int(by ' + bytes) < 0 ↔ ¬ 0 ≤ b2i(by)

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' ≠ [];