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Proof step 9 from proof and-ones (previous step)

∀ bits1, bits0. # bits1 < # bits → true → bits1 = [] ∨ bitsand(bits1, false ' + bits0) = false ' + bits0 ∨ false ∈ bits1 ∨ false ∈ bits0 ⊦ bits = bits.first ', bits0 = [], bits = [], bitsand(bits.butlast, false ' + bits0.butlast) + (bits0.last ∧ bits.last)' = false ' + bits0, false ∈ bits0, false ∈ bits

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
bits = boolvar + bits2, ∀ bits1, bits0. # bits1 < # bits → true → bits1 = [] ∨ bitsand(bits1, false ' + bits0) = false ' + bits0 ∨ false ∈ bits1 ∨ false ∈ bits0 ⊦ bits = boolvar ', bits0 = [], bits = [], bitsand(bits.butlast, false ' + bits0.butlast) + (bits0.last ∧ bits.last)' = false ' + bits0, false ∈ bits0, false ∈ bits

Rule argument

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