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Proof step 31 from proof int2bits-lslen-07 (previous step)

¬(0 ≤ i -1 ∧ 2 ^ i -1 ≤ j + 1 ∧ j + 1 < 2 ^ (i -1 + 1)), j < 2 ^ i * 2, i→n(i + 2) + 1 < # int2bits(j + 1), 2 ^ i ≤ j, 0 < j, 0 ≤ i, j + 1 < 2 ^ i ⊦ 0 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
j + 1 < 2 ^ i, j + 1 < 2 ^ (i - 1), j < 2 ^ i * 2, i→n(i + 2) + 1 < # int2bits(j + 1), 2 ^ i ≤ j, 0 < j, 0 < i ⊦ 0 = j, 0 = i

Rule argument

Additional information

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