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Proof step 1 from proof bits2int-neg (conclusion)

⊦ bits ≠ [] → (bytes2int(bits2bytes(bits)) < 0 ↔ bits.first)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bytes2int(bits2bytes(bits)) < 0 ⊦ bits = [], bits.first

Rule argument

Additional information

Used simplifier rules: bits2int-pos: ⊦ bits ≠ [] → (0 ≤ bytes2int(bits2bytes(bits)) ↔ ¬ bits.first);% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;