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Proof step 23 from proof int2bytes-rec255-neg (previous step)

# int2bytes(i % 256) ≥ 2, i < ~128, i < 0 ⊦ lastn(2, int2bytes(i % 256)).rest = int2bytes(i % 256).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
2 ≤ # int2bytes(i % 256), i < 0, i < ~128 ⊦ lastn(2, int2bytes(i % 256)).rest = int2bytes(i % 256).last '

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ n ≥ m ↔ m ≤ n;