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Proof step 5 from proof bitsor-bi-zeroes (previous step)

⊦ bits = [], bits0 = [], bits2int(bitsor(bits, bits0) + bits1) = bits2int(bits0 + bits1), true ∈ bits, bits0.first

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
bits0 = boolvar + bits2 ⊦ bits = [], bits0 = [], bits2int(bitsor(bits, bits0) + bits1) = bits2int(bits0 + bits1), true ∈ bits, boolvar

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar, bits2] 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: bits0, boolvar, bits2