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Proof step 5 from proof bits2int-lasteq (previous step)

⊦ bits = [], bits0 = [], bits2int(bits + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2int(bits) = bits2int(bits0), boolvar ∧ ¬ bits.first

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
bits = boolvar0 + bits1 ⊦ bits = [], bits0 = [], bits2int(bits + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2int(bits) = bits2int(bits0), boolvar ∧ ¬ boolvar0

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar0, bits1] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar0, bits1