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Proof step 1 from proof bits2bits-2pot-01 (conclusion)

⊦ int2bits(bits2int(true ' + mklist(false, n))) = true ' + mklist(false, n)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);%bits2uint-2pot-02: ⊦ bits2uint(true ' + mklist(false, n)) = 2 ^ n→i(n);%int2bits-2pot-01: 0 ≤ i ⊦ int2bits(~(2 ^ i)) = true ' + mklist(false, i→n(i));% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ i→n(n→i(n)) = n;% ⊦ 0 ≤ n→i(n);% ⊦ (false ' + mklist(true, n))+1 = true ' + mklist(false, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(mklist(boolvar, n)) = mklist(¬ boolvar, n);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ boolvar + bits = boolvar ' + bits;