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Proof step 6 from proof uint2bits-falsein (previous step)

bitsinvert(uint2bits(i)) = bits + boolvar ' ∧ bitsinvert(uint2bits(i)) ≠ [], 1 < i, 0 < i, 0 ≤ i ⊦ boolvar, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bitsinvert(uint2bits(i)) = bits + false ', 0 ≤ i, 0 < i, 1 < i ⊦ false ∈ bits

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];%uint2bits-onein: ⊦ 0 < j → true ∈ uint2bits(j);% ⊦ false ∈ bitsinvert(bits) ↔ true ∈ bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ bitsinvert(bits) = [] ↔ bits = [];