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Proof step 4 from proof fix-bits2bytes (previous step)

¬(# bits2bytes(bits) < n ∧ bytes2int(bits2bytes(bits)) < 0), # bits ≤ 8 * (n - 1) ⊦ bits = [], n = 0, fix(n, bits2bytes(bits)) = bits2bytes(fillsgn(n * 8, bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# bits ≤ 8 * (n - 1) ⊦ bits = [], n = 0, fix(n, bits2bytes(bits)) = bits2bytes(fillsgn(n * 8, bits)), ¬ n ≤ # bits2bytes(bits) ∧ bits.first

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits ≠ [] → (bytes2int(bits2bytes(bits)) < 0 ↔ bits.first);% ⊦ ¬ m < n ↔ n ≤ m;