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Proof step 17 from proof bits2int-lasteq (previous step)

¬ (bits0 + false ').first ∧ bits0 + false ' ≠ [] ⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1))+1) * 2 + bits2uint(bits0 + false ') = 0 ↔ bits2uint((false ' + bitsinvert(bits1))+1) + bits2int(bits0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1))+1) + bits2uint(bits0) = 0 ↔ bits2uint((false ' + bitsinvert(bits1))+1) + bits2int(bits0) = 0, bits0.first

Rule argument

Additional information

Used simplifier rules: bits2uint-zero: ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ bits + boolvar ' ≠ [];% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;