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Proof step 1 from proof bytes2bits-nls (conclusion)

bytes ≠ [] ⊦ ¬ 2 ^ (8 * n→i(# bytes)) + bits2int(bytes2bits(bytes)) < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
2 ^ (8 * n→i(# bytes)) + bits2int(bytes2bits(bytes)) < 0 ⊦ bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)