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Proof step 1 from proof out-test-06 (conclusion)

⊦ b2i(i2b(~512)) = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: num-i2b-neg: i < ~128 ⊦ i2b((* i)) = i2b(bits2int(lastn(8, int2bits(i))));%byte: ⊦ b2i(i2b(0)) = 0;% ⊦ bits2uint(false ') = 0;% ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ [] + bits = bits;% ⊦ lastn(0, bits) = [];%0 < n ⊦ lastn((* n), bits + boolvar ') = lastn(n -1, bits) + boolvar ';% ⊦ sgn(true ' + false ' + bits) = true ' + false ' + bits;% ⊦ sgn(boolvar ' + boolvar ' + bits) = sgn(boolvar ' + bits);% ⊦ (bits + false ')+1 = bits + true;% ⊦ (bits + true ')+1 = bits +1 + false;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bitsinvert(false ') = true ';% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ uint2bits(1) = false + true;%1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);%i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);