Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/3-matchbox)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

1.1 Dependency Pair Transformation

1.1.1 Reduction Pair Processor

1.1.1.1 Monotonic Reduction Pair Processor

1.1.1.1.1 Reduction Pair Processor

1.1.1.1.1.1 Reduction Pair Processor

1.1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  c#(b(a(x1)))	→	a#(b(c(x1)))	(17)
  a#(a(b(c(x1))))	→	c#(b(a(a(x1))))	(9)

  1.1.1.1.1.1.1.1 Reduction Pair Processor

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

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

  the pair

  c#(b(a(x1)))

  →

  a#(b(c(x1)))

  (17)

  could be deleted.

  1.1.1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

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

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

  →

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

  (9)

  could be deleted.

  1.1.1.1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.