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Proof step 23 from proof and-true (previous step)

bitsand(bits.butlast, bits0.butlast) = [], bits0.last, bits.last ⊦ bits0 = bits0.first ', bits = bits.first ', bits = [], bits0 = [], bits = true '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
bits0 = boolvar + bits1, bitsand(bits.butlast, bits0.butlast) = [], bits0.last, bits.last ⊦ bits0 = boolvar ', bits = bits.first ', bits = [], bits0 = [], bits = true '

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar, 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: bits0, boolvar, bits1