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

⊦ bits2int(sublist(0, # fillsgn(64, int2bits(0)), mklist(false, i→n(band(j, 63))) + fillsgn(64, int2bits(0)))) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits2int (sublist (0, 64, mklist(false, i→n(band(j, 63))) + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ')) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n);% ⊦ int2bits(0) = false ';