Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z031)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

{b(☐), c(☐), a(☐)}

1.1 Semantic Labeling

Root-labeling is applied.

1.1.1 Rule Removal

1.1.1.1 Dependency Pair Transformation

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  ab#(bc(cb(bc(ca(aa(aa(x1)))))))	→	ab#(bc(cb(bc(cb(bc(ca(x1)))))))	(25)
  ab#(bc(cb(bc(ca(aa(aa(x1)))))))	→	cb#(bc(cb(bc(ca(x1)))))	(26)
  cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	ab#(bc(cb(bc(cb(bc(cb(x1)))))))	(14)
  cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(bc(cb(bc(cb(x1)))))	(15)
  cb#(bc(cb(bc(ca(aa(aa(x1)))))))	→	ab#(bc(cb(bc(cb(bc(ca(x1)))))))	(18)
  cb#(bc(cb(bc(ca(aa(aa(x1)))))))	→	cb#(bc(cb(bc(ca(x1)))))	(19)
  cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(bc(cb(x1)))	(16)
  cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(x1)	(17)

  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

  [ab#(x1)]	=	-∞ -∞ -∞ + 0 0 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [bc(x1)]	=	-∞ -∞ -∞ + 0 0 0 -∞ 0 -∞ -∞ -∞ 0 · x1
  [cb(x1)]	=	-∞ -∞ -∞ + 0 0 0 -∞ -∞ 0 -∞ 0 -∞ · x1
  [ca(x1)]	=	0 0 -∞ + 0 -∞ -∞ 0 -∞ -∞ -∞ -∞ -∞ · x1
  [aa(x1)]	=	0 -∞ -∞ + -∞ 0 0 0 0 -∞ 0 0 -∞ · x1
  [cb#(x1)]	=	0 -∞ -∞ + 0 0 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [ab(x1)]	=	0 0 0 + 0 0 1 0 0 1 1 1 0 · x1

  the pair

  cb#(bc(cb(bc(ca(aa(ab(x1)))))))

  →

  ab#(bc(cb(bc(cb(bc(cb(x1)))))))

  (14)

  could be deleted.

  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

  [ab#(x1)]	=	0 -∞ -∞ + -∞ 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [bc(x1)]	=	-∞ -∞ -∞ + 0 -∞ 0 -∞ -∞ 0 -∞ 0 -∞ · x1
  [cb(x1)]	=	-∞ -∞ -∞ + 0 0 0 -∞ 0 -∞ -∞ -∞ 0 · x1
  [ca(x1)]	=	0 0 -∞ + -∞ 0 0 1 0 0 0 -∞ 0 · x1
  [aa(x1)]	=	1 -∞ 0 + 0 -∞ 0 -∞ 0 -∞ -∞ -∞ 0 · x1
  [cb#(x1)]	=	0 -∞ -∞ + 0 0 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1
  [ab(x1)]	=	1 -∞ 0 + -∞ -∞ -∞ 0 0 0 -∞ -∞ -∞ · x1

  the pairs

  ab#(bc(cb(bc(ca(aa(aa(x1)))))))	→	ab#(bc(cb(bc(cb(bc(ca(x1)))))))	(25)
  cb#(bc(cb(bc(ca(aa(aa(x1)))))))	→	ab#(bc(cb(bc(cb(bc(ca(x1)))))))	(18)

  could be deleted.

  1.1.1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    cb#(bc(cb(bc(ca(aa(aa(x1)))))))	→	cb#(bc(cb(bc(ca(x1)))))	(19)
    cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(bc(cb(bc(cb(x1)))))	(15)
    cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(bc(cb(x1)))	(16)
    cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(x1)	(17)

    1.1.1.1.1.1.1.1.1 Reduction Pair Processor

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

    [cb#(x1)]	=	0 -∞ -∞ + -∞ 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x1
    [bc(x1)]	=	0 0 0 + 0 0 0 0 0 -∞ 0 -∞ 0 · x1
    [cb(x1)]	=	0 0 0 + -∞ -∞ -∞ -∞ -∞ 0 -∞ 0 -∞ · x1
    [ca(x1)]	=	0 0 0 + -∞ -∞ -∞ -∞ -∞ -∞ 0 -∞ 0 · x1
    [aa(x1)]	=	1 0 0 + -∞ 0 0 0 -∞ 0 -∞ -∞ 0 · x1
    [ab(x1)]	=	0 0 0 + 0 1 0 0 0 1 0 0 0 · x1

    the pair

    cb#(bc(cb(bc(ca(aa(ab(x1)))))))

    →

    cb#(bc(cb(x1)))

    (16)

    could be deleted.

    1.1.1.1.1.1.1.1.1.1 Reduction Pair Processor

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

    [cb#(x1)]	=	0 -∞ -∞ + -∞ 0 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1
    [bc(x1)]	=	0 0 0 + 0 0 0 -∞ 0 -∞ -∞ -∞ 0 · x1
    [cb(x1)]	=	0 -∞ 0 + 0 0 0 -∞ -∞ 0 -∞ 0 -∞ · x1
    [ca(x1)]	=	1 0 1 + 0 -∞ 0 -∞ 0 -∞ 0 1 0 · x1
    [aa(x1)]	=	0 0 0 + -∞ 0 0 0 -∞ -∞ 0 0 0 · x1
    [ab(x1)]	=	0 -∞ 0 + -∞ 0 0 -∞ 0 -∞ -∞ 0 -∞ · x1

    the pairs

    cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(bc(cb(bc(cb(x1)))))	(15)
    cb#(bc(cb(bc(ca(aa(ab(x1)))))))	→	cb#(x1)	(17)

    could be deleted.

    1.1.1.1.1.1.1.1.1.1.1 Switch to Innermost Termination

    The TRS does not have overlaps with the pairs and is locally confluent:

    20

    Hence, it suffices to show innermost termination in the following.

    1.1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

    We restrict the rewrite rules to the following usable rules of the DP problem.

    There are no rules.

    1.1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

    We restrict the innermost strategy to the following left hand sides.

    cb(bc(cb(bc(ca(aa(ab(x0)))))))

    cb(bc(cb(bc(ca(aa(aa(x0)))))))

    1.1.1.1.1.1.1.1.1.1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [cb#(x1)]	=	1 · x1
    [bc(x1)]	=	1 · x1
    [cb(x1)]	=	1 · x1
    [ca(x1)]	=	1 · x1
    [aa(x1)]	=	1 + 1 · x1

    the pair

    cb#(bc(cb(bc(ca(aa(aa(x1)))))))

    →

    cb#(bc(cb(bc(ca(x1)))))

    (19)

    could be deleted.

    1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.