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Proof step 36 from proof int2bits-lslen-07 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
2 ^ i * 2 = j + 1, i→n(i + 2) + 1 < # int2bits(j + 1), 2 ^ i ≤ j, 0 ≤ i, 0 < j ⊦ 0 = j

Rule argument

Additional information

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