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Proof step 2 from proof bitsand-zeros (previous step)

# bits ≤ # bits0 ∨ bits0.first ⊦ bits = [], bits0 = [], bitsand(bits, bits0) = bits, true ∈ bits

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
bits0 = boolvar + bits1, # bits ≤ # bits0 ∨ boolvar ⊦ bits = [], bits0 = [], bitsand(bits, bits0) = bits, true ∈ bits

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