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Proof step 1 from proof int2bits-true (conclusion)

⊦ int2bits(j) = true ' ↔ j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
int2bits(j) = true ' ⊦ j = ~1

Rule argument

Additional information

Used simplifier rules: int2bits-neg-num: i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);%uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ sgn(true ' + true ') = true ';% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + false ')+1 = bits + true;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bitsinvert(true ') = false ';% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';