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Proof step 20 from proof bits2int-sign (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
bits0.first ⊦ bits2int(bits0 + false ' + bits1) = bits2uint(bits1), bits0 = [], true ∈ bits0

Rule argument

Additional information

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