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Proof step 6 from proof int2bits-len-range (previous step)

i→n(i + 1) < # int2bits(j), 0 ≤ 2 ^ i + j, j < 2 ^ i, j ≠ 0 → 0 ≤ i0 ∧ 2 ^ i0 ≤ abs(j) ∧ abs(j) < 2 ^ (i0 + 1) ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 < 2 ^ i0, i→n(i + 1) < # int2bits(j), 0 ≤ 2 ^ i + j, j < 2 ^ i, 0 ≤ i0, 2 ^ i0 ≤ abs(j), abs(j) < 2 ^ i0 * 2 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;% ⊦ j < (* i1) * j ↔ 0 < (i1 - 1) * j;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;