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Proof step 9 from proof int2bits-len-range (previous step)

¬ j < 0, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 ≤ 2 ^ i + j, j < 2 ^ i, 0 ≤ i0, 2 ^ i0 ≤ j, j < 2 ^ i0 * 2, 1 < 2 ^ i, 0 < i, 0 ≤ 2 ^ i, 0 < 2 ^ i, 0 < j, i0 < i, 0 < 2 ^ i0, i→n(i) + 1 < # int2bits(j) ⊦

Rule argument

Additional information

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