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Proof step 6 from proof bits2int-ones-02 (previous step)

(bits + bits0).first ∧ bits + bits0 ≠ [] ⊦ bits = [], bits2uint((false ' + bitsinvert(bits0))+1) + ~ bits2uint(bitsinvert(bits + bits0) +1) = 0, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bits.first ⊦ bits + bits0 = [], bits = [], bits2uint((false ' + bitsinvert(bits0))+1) = bits2uint((bitsinvert(bits) + bitsinvert(bits0))+1), false ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);