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Proof step 14 from proof int2bits-lslen-07 (previous step)

¬(0 ≤ i ∧ ~(2 ^ (i + 1)) ≤ j + 1 ∧ j + 1 < ~(2 ^ i)), j < 0, j + 2 ^ i < 0, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j, 3 + i→n(i) < # int2bits(j + 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j + 2 ^ i = ~1, 3 + i→n(i) < # int2bits(j + 1), 0 ≤ 2 ^ i * 2 + j, 0 ≤ i, j < 0 ⊦

Rule argument

Additional information

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