Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/teparla1)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  f#(f(y,z),f(x,f(a,x)))

  →

  f#(f(f(a,z),f(x,a)),f(a,y))

  (2)

  1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [f(x1, x2)]	=	1 · x1 + 1 · x2
  [a]	=	0
  [f#(x1, x2)]	=	1 · x1 + 1 · x2

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  f#(f(f(a,z1),f(z2,a)),f(a,f(a,a)))

  →

  f#(f(f(a,f(z2,a)),f(a,a)),f(a,f(a,z1)))

  (7)

  1.1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  f#(f(f(a,f(z1,a)),f(a,a)),f(a,f(a,a)))

  →

  f#(f(f(a,f(a,a)),f(a,a)),f(a,f(a,f(z1,a))))

  (8)

  1.1.1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  There are no rules.

  1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.