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Proof step 18 from proof int2bytes-byte-neg (previous step)

# restn(1, int2bits(i)) ≤ 8, ~256 < i, i < ~128, i < 0 ⊦ i2b(bits2int(restn(1, int2bits(i)))) ' = i2b(i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
~256 < i, i < ~128, i < 0, # int2bits(i) ≤ 9 ⊦ i2b(bits2int(restn(1, int2bits(i)))) = i2b(i)

Rule argument

Additional information

Used simplifier rules: 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) ≠ [];% ⊦ by ' = by0 ' ↔ by = by0;