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Proof step 9 from proof shiftright-minus-one (previous step)

1 ≠ 0 ⊦ bits2int((mklist(true, i→n(j)) + true ').first ' + firstn(1 -1, (mklist(true, i→n(j)) + true ').rest)) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ bits2int((mklist(true, i→n(j)) + true ').first ') = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ firstn(0, bits) = [];