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Proof step 1 from proof bytes2bytes-first-neg (conclusion)

⊦ b2i(int2bytes(bytes2int(by ' + bytes)).first) < 0 ↔ b2i(by) < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ b2i(int2bytes(bytes2int(by ' + bytes)).first) < 0 ↔ ¬ 0 ≤ b2i(by)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;