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Proof step 4 from proof int2bits-rec-2pot-neg-01 (previous step)

2 ^ i ≠ 0, ∃ j0. j = j0 * 2 ^ i, j + 2 ^ i < i, j + 2 ^ i ≤ 0, 0 < i ⊦ int2bits(j) = int2bits(j / 2 ^ i) + mklist(false, i→n(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i, 2 ^ i * (j0 + 1) ≤ 0, 2 ^ i * (j0 + 1) < i ⊦ int2bits(j0 * 2 ^ i) = int2bits(j0) + mklist(false, i→n(i))

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j + i = i * (j + 1);% ⊦ 0 < j → (i ^ j = 0 ↔ i = 0);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;%j ≠ 0 ⊦ (i * j) / j = i;