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Proof step 26 from proof int2bits-lslen-07 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
0 ≤ i, 0 < j, 2 ^ i ≤ j, i→n(i + 2) + 1 < # int2bits(j + 1), j < 2 ^ i * 2 ⊦ 0 = j, ¬ j + 1 < 2 ^ i ∧ 2 ^ i * 2 ≠ j + 1

Rule argument

Additional information

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