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Proof step 22 from proof bits2int-sign (previous step)

¬ (bits0 + false ' + bits1).first ∧ bits0 + false ' + bits1 ≠ [] ⊦ bits2uint((bits0 + false ' + bits1).rest) = bits2uint(bits1), bits0 = [], true ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ bits0 = [], bits2uint(bits0.rest + false ' + bits1) = bits2uint(bits1), true ∈ bits0, bits0.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;