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Proof step 33 from proof lastn-fill-sgn (previous step)

# bits0 = # fillsgn(# bits0, sgn(bits0)), bits0.first, true ∈ bits1, # sgn(bits0) < # bits1 + # bits0, # sgn(bits0) < # bits0, 1 + # sgn(bits0) < # bits1 + # bits0 ⊦ bits0 = [], bits1 + bits0 = [], bits1 = [], fillsgn(# bits0, sgn(bits0)) = bits0, false ∈ bits1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 34:
bits0 = boolvar + bits, # bits0 = # fillsgn(# bits0, sgn(bits0)), boolvar, true ∈ bits1, # sgn(bits0) < # bits1 + # bits0, # sgn(bits0) < # bits0, 1 + # sgn(bits0) < # bits1 + # bits0 ⊦ bits0 = [], bits1 + bits0 = [], bits1 = [], fillsgn(# bits0, sgn(bits0)) = bits0, false ∈ bits1

Rule argument

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