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

¬ (boolvar ' + bits1).first ∧ boolvar ' + bits1 ≠ [] ⊦ bits0 = [], bits2int(bits0 + boolvar ' + bits1) = bits2uint((boolvar ' + bits1).rest), (¬ boolvar) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
⊦ bits2int(bits0 + false ' + bits1) = bits2uint(bits1), bits0 = [], true ∈ bits0

Rule argument

Additional information

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