Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup05)

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 2 components.

• The 1^st component contains the pair

  s#(s(a(a(x1))))	→	s#(x1)	(12)
  s#(s(a(a(x1))))	→	s#(s(x1))	(11)
  s#(s(b(b(a(a(b(b(x1))))))))	→	s#(s(b(b(a(a(x1))))))	(15)

  1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pairs

  s#(s(a(a(x1))))	→	s#(x1)	(12)
  s#(s(a(a(x1))))	→	s#(s(x1))	(11)
  s#(s(b(b(a(a(b(b(x1))))))))	→	s#(s(b(b(a(a(x1))))))	(15)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(a(a(b(b(a(a(x1)))))))	(22)
  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(b(a(a(b(b(a(a(x1))))))))	(21)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(a(b(b(a(a(x1))))))	(23)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(a(b(b(a(a(b(b(x1))))))))	(29)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(b(b(a(a(b(b(x1)))))))	(30)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(b(a(a(b(b(x1))))))	(31)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(b(b(a(a(x1)))))	(24)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(a(a(b(b(x1)))))	(32)
  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(b(a(a(x1))))	(25)
  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(a(a(x1)))	(26)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(a(x1))	(27)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(a(b(b(x1))))	(33)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(b(b(x1)))	(34)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(b(x1))	(35)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(x1)	(28)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(x1)	(36)

  1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [b(x1)]	=	1 · x1
  [a(x1)]	=	1 · x1
  [b#(x1)]	=	1 · x1
  [a#(x1)]	=	1 · x1

  together with the usable rules

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

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

  1.1.1.2.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pairs

  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(a(a(b(b(a(a(x1)))))))	(22)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(a(b(b(a(a(x1))))))	(23)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(b(b(a(a(b(b(x1)))))))	(30)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(b(a(a(b(b(x1))))))	(31)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(b(b(a(a(x1)))))	(24)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(a(a(b(b(x1)))))	(32)
  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(b(a(a(x1))))	(25)
  b#(b(b(b(a(a(b(b(x1))))))))	→	b#(a(a(x1)))	(26)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(a(x1))	(27)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(a(b(b(x1))))	(33)
  a#(a(a(a(b(b(a(a(x1))))))))	→	a#(b(b(x1)))	(34)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(b(x1))	(35)
  b#(b(b(b(a(a(b(b(x1))))))))	→	a#(x1)	(28)
  a#(a(a(a(b(b(a(a(x1))))))))	→	b#(x1)	(36)

  could be deleted.

  1.1.1.2.1.1 Dependency Graph Processor

  The dependency pairs are split into 2 components.

  • The 1^st component contains the pair

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

    →

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

    (21)

    1.1.1.2.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

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

    the pair

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

    →

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

    (21)

    could be deleted.

    1.1.1.2.1.1.1.1 P is empty

    There are no pairs anymore.

  • The 2^nd component contains the pair

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

    →

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

    (29)

    1.1.1.2.1.1.2 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

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

    the pair

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

    →

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

    (29)

    could be deleted.

    1.1.1.2.1.1.2.1 P is empty

    There are no pairs anymore.