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Proof step 15 from proof int2bits-lslen-08 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
0 = 2 ^ i * 2 + j, 0 ≤ i, j + 2 ^ i < 0, j < 0, i→n(i + 2) + 1 < # int2bits(j - 1) ⊦

Rule argument

Additional information

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