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Proof step 24 from proof bits2int-lasteq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
⊦ bits0 = [], bits2uint(bits1) = bits2uint(bits0) ↔ bits2uint(bits1) = bits2int(bits0), 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 ' ≠ [];% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;