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Proof step 7 from proof bits2int-zeros-01 (previous step)

¬ (bits + false ' + bits0).first ∧ bits + false ' + bits0 ≠ [] ⊦ bits = [], bits2uint((bits + false ' + bits0).rest) = bits2uint(bits0), true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ bits = [], bits2uint(bits.rest + false ' + bits0) = bits2uint(bits0), true ∈ bits, bits.first

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ bits + boolvar ' + bits0 ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits + bits0).rest = bits.rest + bits0;