Certification Problem

Input (TPDB SRS_Standard/Zantema_06/12)

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

  l#(r(a(a(x1))))	→	a#(a(l(c(c(c(r(x1)))))))	(10)
  a#(l(x1))	→	l#(a(x1))	(5)
  l#(r(a(a(x1))))	→	a#(l(c(c(c(r(x1))))))	(11)
  a#(l(x1))	→	a#(x1)	(6)
  a#(c(x1))	→	c#(a(x1))	(7)
  c#(a(r(x1)))	→	a#(x1)	(9)
  a#(c(x1))	→	a#(x1)	(8)

  1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [l#(x1)]	=	1 + 1 · x1
  [r(x1)]	=	1 + 1 · x1
  [a(x1)]	=	1 · x1
  [a#(x1)]	=	1 + 1 · x1
  [l(x1)]	=	1 · x1
  [c(x1)]	=	1 · x1
  [c#(x1)]	=	1 · x1

  the pair

  a#(c(x1))

  →

  c#(a(x1))

  (7)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    a#(l(x1))	→	l#(a(x1))	(5)
    l#(r(a(a(x1))))	→	a#(a(l(c(c(c(r(x1)))))))	(10)
    a#(l(x1))	→	a#(x1)	(6)
    a#(c(x1))	→	a#(x1)	(8)
    l#(r(a(a(x1))))	→	a#(l(c(c(c(r(x1))))))	(11)

    1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [a#(x1)]	=	1 + 1 · x1
    [l(x1)]	=	1 + 1 · x1
    [l#(x1)]	=	1 + 1 · x1
    [a(x1)]	=	1 + 1 · x1
    [r(x1)]	=	1 + 1 · x1
    [c(x1)]	=	1 · x1

    the pairs

    a#(l(x1))	→	a#(x1)	(6)
    l#(r(a(a(x1))))	→	a#(l(c(c(c(r(x1))))))	(11)

    could be deleted.

    1.1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

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

    the pair

    a#(c(x1))

    →

    a#(x1)

    (8)

    could be deleted.

    1.1.1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the rationals with delta = 3/32

    [a#(x1)]	=	0 + 1 · x1
    [l(x1)]	=	0 + 2 · x1
    [l#(x1)]	=	0 + 1/2 · x1
    [a(x1)]	=	0 + 4 · x1
    [r(x1)]	=	1/4 + 0 · x1
    [c(x1)]	=	0 + 1/4 · x1

    the pair

    l#(r(a(a(x1))))

    →

    a#(a(l(c(c(c(r(x1)))))))

    (10)

    could be deleted.

    1.1.1.1.1.1.1.1 Monotonic Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [a(x1)]	=	1 · x1
    [l(x1)]	=	2 + 1 · x1
    [c(x1)]	=	1 · x1
    [r(x1)]	=	1 · x1
    [a#(x1)]	=	3 + 2 · x1
    [l#(x1)]	=	2 · x1

    the pair

    a#(l(x1))

    →

    l#(a(x1))

    (5)

    and no rules could be deleted.

    1.1.1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.