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Proof step 1 from proof bytes2int-pos (conclusion)

⊦ 0 ≤ bytes2int(bytes) ↔ bytes = [] ∨ 0 ≤ b2i(bytes.first)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ 0 ≤ bytes2int(bytes) ↔ bytes = [] ∨ ¬ b2i(bytes.first) < 0

Rule argument

Additional information

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