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

# int2bits(j) % 8 ≠ 0, j + 2 ^ (8 * i) ≤ 0, 0 ≤ 2 ^ (8 * (i + 1) - 1) + j, 0 ≤ i, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1) ⊦ # int2bits(j) / 8 + 1 = i→n(i) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
j + 2 ^ (8 * i) ≤ 0, 0 ≤ 2 ^ (8 * (i + 1) - 1) + j, 0 ≤ i, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1) ⊦ # int2bits(j) % 8 = 0, # int2bits(j) / 8 = i→n(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;