Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/bababbababa-abababbababba.srs)

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#(b(a(b(a(b(b(a(b(a(b(x1)))))))))))	→	a#(b(a(x1)))	(7)
  a#(b(a(b(a(b(b(a(b(a(b(x1)))))))))))	→	a#(b(a(b(a(x1)))))	(6)
  a#(b(a(b(a(b(b(a(b(a(b(x1)))))))))))	→	a#(x1)	(8)

  1.1.1.1 Semantic Labeling Processor

  The following interpretations form a model of the rules.

  As carrier we take the set {0,1}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 2):

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

  We obtain the set of labeled pairs

  a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(b1(a0(x1)))	(9)
  a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(b1(a0(b1(a0(x1)))))	(10)
  a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(b0(a1(b0(a1(x1)))))	(11)
  a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(b0(a1(x1)))	(12)
  a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(x1)	(13)
  a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(x1)	(14)

  and the set of labeled rules:

  a0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a0(b1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(a0(x1)))))))))))))	(15)
  a1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a1(b0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(a1(x1)))))))))))))	(16)

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 2 components.

  • The 1^st component contains the pair

    a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(b1(a0(b1(a0(x1)))))	(10)
    a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(b1(a0(x1)))	(9)
    a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(x1)	(13)

    1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

    [a0(x1)]	=	1 · x1
    [b1(x1)]	=	1 · x1
    [b0(x1)]	=	1 · x1
    [a1(x1)]	=	1 · x1
    [a#0(x1)]	=	1 · x1

    together with the usable rule

    a0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))

    →

    a0(b1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(a0(x1)))))))))))))

    (15)

    (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

    1.1.1.1.1.1.1 Monotonic Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [a0(x1)]	=	1 · x1
    [b1(x1)]	=	1 · x1
    [b0(x1)]	=	1 · x1
    [a1(x1)]	=	1 + 1 · x1
    [a#0(x1)]	=	1 · x1

    the pairs

    a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(b1(a0(b1(a0(x1)))))	(10)
    a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(b1(a0(x1)))	(9)
    a#0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(x1)))))))))))	→	a#0(x1)	(13)

    and no rules could be deleted.

    1.1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.

  • The 2^nd component contains the pair

    a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(b0(a1(x1)))	(12)
    a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(b0(a1(b0(a1(x1)))))	(11)
    a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(x1)	(14)

    1.1.1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

    [a1(x1)]	=	1 · x1
    [b0(x1)]	=	1 · x1
    [b1(x1)]	=	1 · x1
    [a0(x1)]	=	1 · x1
    [a#1(x1)]	=	1 · x1

    together with the usable rule

    a1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))

    →

    a1(b0(b1(a0(b1(a0(b1(b0(a1(b0(a1(b0(a1(x1)))))))))))))

    (16)

    (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

    1.1.1.1.1.2.1 Monotonic Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [a1(x1)]	=	1 · x1
    [b0(x1)]	=	1 · x1
    [b1(x1)]	=	1 · x1
    [a0(x1)]	=	1 + 1 · x1
    [a#1(x1)]	=	1 · x1

    the pairs

    a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(b0(a1(x1)))	(12)
    a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(b0(a1(b0(a1(x1)))))	(11)
    a#1(b0(a1(b0(a1(b0(b1(a0(b1(a0(b1(x1)))))))))))	→	a#1(x1)	(14)

    and no rules could be deleted.

    1.1.1.1.1.2.1.1 P is empty

    There are no pairs anymore.