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Proof step 15 from proof bits2int-lasteq (previous step)

¬(¬ (bits0 + false ').first ∧ bits0 + false ' ≠ []) ⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1))+1) * 2 + bits2int(bits0 + false ') = 0 ↔ bits2uint((false ' + bitsinvert(bits1))+1) + bits2int(bits0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
bits0.first ⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1))+1) * 2 + bits2int(bits0 + false ') = 0 ↔ bits2uint((false ' + bitsinvert(bits1))+1) + bits2int(bits0) = 0

Rule argument

Additional information

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