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Proof step 1 from proof int2bits-011x (conclusion)

⊦ ¬ false ∈ int2bits(4294967295).rest

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: int2bits-pos-num: 0 ≤ i ⊦ int2bits((* i)) = uint2bits(i);%uint2bits-pos-num: 1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);%uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';