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Proof step 6 from proof and-first (previous step)

⊦ bitsand(boolvar ' + bits1, boolvar0 ' + bits).first ↔ boolvar ∧ boolvar0

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] → [bitsand(boolvar ' + bits1, boolvar0 ' + bits), boolvar1, bits0] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: and-notempty-06: ⊦ bitsand(boolvar ' + bits, boolvar0 ' + bits0) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bitsand(boolvar ' + bits1, boolvar0 ' + bits), boolvar1, bits0