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Proof step 4 from proof bytes2bytes-fillsgn (previous step)

¬ # int2bytes(bytes2int(bytes)) ≥ # bytes ⊦ fillsgn(# bytes, int2bytes(bytes2int(bytes))) = bytes, bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# int2bytes(bytes2int(bytes)) < # bytes ⊦ bytes = [], fillsgn(# bytes, int2bytes(bytes2int(bytes))) = bytes

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ n ≥ m ↔ m ≤ n;