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Proof step 4 from proof shiftright-neg-mone (previous step)

0 ≤ 2 ^ (abs(j) - 1) + i, i < 0 ⊦ bits2int(sublist(0, # int2bits(i), mklist(int2bits(i).first, i→n(j)) + int2bits(i))) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ 2 ^ (abs(j) - 1) + i, i < 0 ⊦ bits2int(sublist(0, # int2bits(i), mklist(true, i→n(j)) + int2bits(i))) = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ j < 0 → int2bits(j).first;