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Proof step 22 from proof fillsgn-short (previous step)

# bytes = 1, fillsgn(2, int2bytes(bytes2int(bytes))) = bytes, # bytes ≤ # fillsgn(# bytes, int2bytes(bytes2int(bytes))), # bytes < 2 ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
# bytes = 1, fillsgn(2, int2bytes(bytes2int(bytes))) = bytes, 1 ≤ # fillsgn(1, int2bytes(bytes2int(bytes))) ⊦ fillsgn(1, int2bytes(bytes2int(bytes))) = [], bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)