Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z008)

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#(b(a(b(x1))))	→	s#(b(a(x1)))	(12)
  s#(a(x1))	→	s#(x1)	(10)

  1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  together with the usable rules

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

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

  1.1.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pairs

  s#(b(a(b(x1))))	→	s#(b(a(x1)))	(12)
  s#(a(x1))	→	s#(x1)	(10)

  and no rules could be deleted.

  1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  b#(b(a(b(x1))))	→	a#(b(a(x1)))	(16)
  a#(a(b(a(x1))))	→	a#(b(a(b(x1))))	(19)
  a#(a(b(a(x1))))	→	b#(a(b(x1)))	(20)
  b#(b(a(b(x1))))	→	b#(a(b(a(x1))))	(15)
  b#(b(a(b(x1))))	→	b#(a(x1))	(17)
  b#(b(a(b(x1))))	→	a#(x1)	(18)
  a#(a(b(a(x1))))	→	a#(b(x1))	(21)
  a#(a(b(a(x1))))	→	b#(x1)	(22)

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

  together with the usable rules

  b(b(a(b(x1))))	→	b(a(b(a(x1))))	(7)
  a(a(b(a(x1))))	→	a(b(a(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)]	=	3 + 2 · x1
  [b(x1)]	=	1 + 2 · x1
  [a(x1)]	=	1 + 2 · x1
  [a#(x1)]	=	2 + 4 · x1

  the pairs

  b#(b(a(b(x1))))	→	a#(b(a(x1)))	(16)
  a#(a(b(a(x1))))	→	b#(a(b(x1)))	(20)
  b#(b(a(b(x1))))	→	b#(a(x1))	(17)
  b#(b(a(b(x1))))	→	a#(x1)	(18)
  a#(a(b(a(x1))))	→	a#(b(x1))	(21)
  a#(a(b(a(x1))))	→	b#(x1)	(22)

  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

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

    →

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

    (19)

    1.1.1.2.1.1.1 Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

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

    together with the usable rule

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

    →

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

    (7)

    (w.r.t. the implicit argument filter of the reduction pair), the pair

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

    →

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

    (19)

    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

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

    →

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

    (15)

    1.1.1.2.1.1.2 Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

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

    together with the usable rule

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

    →

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

    (8)

    (w.r.t. the implicit argument filter of the reduction pair), the pair

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

    →

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

    (15)

    could be deleted.

    1.1.1.2.1.1.2.1 P is empty

    There are no pairs anymore.