Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/8)

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

  c#(c(c(y)))	→	c#(a(a(c(b(0,y)),0),0))	(5)
  c#(c(c(y)))	→	c#(c(a(a(c(b(0,y)),0),0)))	(4)

  1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [b(x1, x2)]	=	1 · x1 + 1 · x2
  [0]	=	0
  [c(x1)]	=	1 + 2 · x1
  [a(x1, x2)]	=	1 · x1 + 2 · x2
  [c#(x1)]	=	2 · x1

  the pair

  c#(c(c(y)))

  →

  c#(a(a(c(b(0,y)),0),0))

  (5)

  and no rules could be deleted.

  1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 1 to get the following set of pairs

  c#(c(c(y0)))	→	c#(c(b(a(c(b(0,y0)),0),0)))	(11)
  c#(c(c(y0)))	→	c#(c(a(b(c(b(0,y0)),0),0)))	(12)

  1.1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 1 to get the following set of pairs

  c#(c(c(y0)))

  →

  c#(c(b(b(c(b(0,y0)),0),0)))

  (13)

  1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    c#(c(c(y0)))

    →

    c#(c(a(b(c(b(0,y0)),0),0)))

    (12)

    1.1.1.1.1.1.1 Narrowing Processor

    We consider all narrowings of the pair below position 1 to get the following set of pairs

    c#(c(c(y0)))

    →

    c#(c(b(b(c(b(0,y0)),0),0)))

    (13)

    1.1.1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

    [c(x1)]	=	1 · x1
    [b(x1, x2)]	=	1 · x1 + 1 · x2
    [0]	=	0
    [c#(x1)]	=	1 · x1

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

    1.1.1.1.1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

    There are no rules.

    1.1.1.1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.