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Proof step 3 from proof shiftright-abs (previous step)

⊦ bits2int(sublist(0, # int2bits(i), mklist(int2bits(i).first, i→n(j)) + int2bits(i))) = i >> abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(j)) + int2bits(i))) = i >> abs(j)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ int2bits(j).first ↔ j < 0;