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Proof step 7 from proof and-true (previous step)

bitsand(boolvar ' + bits1, bits0) = true ' ⊦ bits0 = [], boolvar ' + bits1 = true ' ∧ bits0 = true ', bits0 ≠ bits0.first ' ∧ bits1 ≠ []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits0, boolvar0, bits] 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, boolvar0, bits