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Proof step 15 from proof int2bytes-byte-neg (previous step)

¬ # restn(1, int2bits(i)) ≤ 8, ~256 < i, i < ~128, i < 0 ⊦ bits2bytes(restn(1, int2bits(i))) = i2b(i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
9 < # int2bits(i), i < 0, i < ~128, ~256 < i ⊦ bits2bytes(restn(1, int2bits(i))) = i2b(i) '

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;%n ≤ # bits ⊦ # restn(n, bits) = # bits - n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ int2bits(j) ≠ [];