Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-98)

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#(b(b(x1)))	→	a#(a(b(a(x1))))	(5)
  a#(a(x1))	→	c#(x1)	(4)
  c#(b(b(x1)))	→	a#(x1)	(7)

  1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

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

  the pair

  c#(b(b(x1)))

  →

  a#(x1)

  (7)

  could be deleted.

  1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

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

  the pair

  a#(a(x1))

  →

  c#(x1)

  (4)

  could be deleted.

  1.1.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pair

  c#(b(b(x1)))

  →

  a#(a(b(a(x1))))

  (5)

  and no rules could be deleted.

  1.1.1.1.1.1 P is empty

  There are no pairs anymore.