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Proof step 1 from proof int2bytes-fillsgn-rec (conclusion)

⊦ 1 < # int2bytes(i) ∧ # int2bytes(i) ≤ n → fillsgn(n, int2bytes(i)) = fillsgn(n -1, int2bytes(i >> 8)) + i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# int2bytes(i) ≤ n ⊦ fillsgn(n, int2bytes(i)) = fillsgn(n - 1, int2bytes(i >> 8)) + i2b(i) ', # int2bytes(i) = 1

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ bytes + by = bytes + by ';% ⊦ n -1 = n - 1;