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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits2int(sublist(0, 1, mklist(false, i→n(j)) + false ')) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar '.first ↔ boolvar;% ⊦ # boolvar ' = 0 +1;% ⊦ int2bits(0) = false ';