Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/7)

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(a(x,y))))	→	c#(c(c(y)))	(5)
  c#(c(c(a(x,y))))	→	c#(c(c(c(y))))	(4)
  c#(c(c(a(x,y))))	→	c#(c(y))	(6)
  c#(c(c(a(x,y))))	→	c#(y)	(7)
  c#(c(b(c(y),0)))	→	c#(c(a(y,0)))	(8)
  c#(c(a(a(y,0),x)))	→	c#(y)	(10)

  1.1.1 Reduction Pair Processor

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

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

  the pairs

  c#(c(c(a(x,y))))	→	c#(c(c(y)))	(5)
  c#(c(c(a(x,y))))	→	c#(c(y))	(6)
  c#(c(c(a(x,y))))	→	c#(y)	(7)
  c#(c(b(c(y),0)))	→	c#(c(a(y,0)))	(8)
  c#(c(a(a(y,0),x)))	→	c#(y)	(10)

  could be deleted.

  1.1.1.1 Reduction Pair Processor

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

  [c#(x1)]	=	3 + 0 · x1
  [c(x1)]	=	-∞ + 1 · x1
  [a(x1, x2)]	=	2 + 0 · x1 + 2 · x2
  [b(x1, x2)]	=	4 + 1 · x1 + -∞ · x2
  [0]	=	0

  the pair

  c#(c(c(a(x,y))))

  →

  c#(c(c(c(y))))

  (4)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.