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Proof step 11 from proof int2bits-lslen-07 (previous step)

0 ≤ i ∧ 0 ≤ 2, (i→n(i) + i→n(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 12:
j < 0, j + 2 ^ i < 0, 0 ≤ i, 0 ≤ 2 ^ i * 2 + j, 3 + i→n(i) < # int2bits(j + 1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;