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Proof step 4 from proof int2bits-less-2pot-01 (previous step)

¬ j < 0, 2 ^ (n→i(# int2bits(j)) - 1) < abs(j) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 ^ (n→i(# int2bits(j)) - 1) < j, 0 ≤ j ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ j < 0 → abs(j) = j;