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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
bits0.first ⊦ bits0 = [], bits2uint(bits1) * 2 = bits2int(bits0 + false ') ↔ bits2uint(bits1) = bits2int(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);