Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-18)

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

  b#(b(y,z),c(a,a,a))	→	c#(z,y,z)	(5)
  c#(y,x,f(z))	→	b#(f(b(z,x)),z)	(6)
  c#(y,x,f(z))	→	b#(z,x)	(8)

  1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 1 with strict dimension 1 over the arctic semiring over the naturals

  [b#(x1, x2)]	=	-∞ + 0 · x1 + -∞ · x2
  [b(x1, x2)]	=	0 + 1 · x1 + 1 · x2
  [c(x1, x2, x3)]	=	1 + -∞ · x1 + 2 · x2 + 2 · x3
  [a]	=	0
  [c#(x1, x2, x3)]	=	0 + -∞ · x1 + 1 · x2 + 1 · x3
  [f(x1)]	=	-∞ + 0 · x1

  the pair

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

  →

  b#(z,x)

  (8)

  could be deleted.

  1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

  [b#(x1, x2)]	=	0 0 + 0 2 0 0 · x1 + 0 0 0 0 · x2
  [b(x1, x2)]	=	0 2 + 0 2 0 1 · x1 + 0 0 2 2 · x2
  [c(x1, x2, x3)]	=	0 2 + 0 0 0 0 · x1 + 0 0 0 0 · x2 + 2 0 2 2 · x3
  [a]	=	0 1
  [c#(x1, x2, x3)]	=	0 0 + 2 3 0 0 · x1 + 0 0 0 0 · x2 + 2 0 0 0 · x3
  [f(x1)]	=	0 0 + 0 2 1 0 · x1

  the pair

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

  →

  c#(z,y,z)

  (5)

  could be deleted.

  1.1.1.1.1 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  c#(y,x,f(z))	→	b#(f(b(z,x)),z)	(6)
  3	>	2

  As there is no critical graph in the transitive closure, there are no infinite chains.