Certification Problem

Input (TPDB SRS_Standard/Trafo_06/hom01)

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 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  a#(d(b(d(b(a(a(x1)))))))	→	a#(c(a(a(x1))))	(8)
  a#(d(b(d(b(a(a(x1)))))))	→	a#(a(c(a(a(x1)))))	(7)
  a#(d(b(d(b(a(a(x1)))))))	→	c#(a(a(x1)))	(9)
  c#(a(a(x1)))	→	a#(a(c(c(x1))))	(10)
  c#(a(a(x1)))	→	a#(c(c(x1)))	(11)
  c#(a(a(x1)))	→	c#(c(x1))	(12)
  c#(a(a(x1)))	→	c#(x1)	(13)

  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

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

  the pairs

  a#(d(b(d(b(a(a(x1)))))))	→	a#(c(a(a(x1))))	(8)
  a#(d(b(d(b(a(a(x1)))))))	→	a#(a(c(a(a(x1)))))	(7)
  a#(d(b(d(b(a(a(x1)))))))	→	c#(a(a(x1)))	(9)

  could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    c#(a(a(x1)))	→	c#(x1)	(13)
    c#(a(a(x1)))	→	c#(c(x1))	(12)

    1.1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

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

    the pairs

    c#(a(a(x1)))	→	c#(x1)	(13)
    c#(a(a(x1)))	→	c#(c(x1))	(12)

    could be deleted.

    1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.