Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x06)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Reduction Pair Processor

1.1.1 Reduction Pair Processor

1.1.1.1 Reduction Pair Processor

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  d#(x1)	→	c#(e(x1))	(14)
  c#(c(x1))	→	d#(b(x1))	(12)

  1.1.1.1.1.1 Reduction Pair Processor

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

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

  the pair

  c#(c(x1))

  →

  d#(b(x1))

  (12)

  could be deleted.

  1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [c#(x1)]	=	1
  [c(x1)]	=	-2 + 2 · x1
  [d(x1)]	=	2 + x1
  [b(x1)]	=	-2 + 2 · x1
  [e(x1)]	=	1 + x1
  [g(x1)]	=	2 + x1
  [f(x1)]	=	1 + x1
  [a(x1)]	=	x1
  [d#(x1)]	=	2 + 2 · x1

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

  d#(x1)

  →

  c#(e(x1))

  (14)

  could be deleted.

  1.1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.