Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr10)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

1.1 Rule Removal

1.1.1 Rule Removal

1.1.1.1 Rule Removal

1.1.1.1.1 Dependency Pair Transformation

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  c#(a(c(x1)))	→	c#(a(a(x1)))	(19)
  c#(a(c(x1)))	→	a#(a(x1))	(18)
  a#(a(a(x1)))	→	a#(c(c(x1)))	(13)
  a#(c(a(x1)))	→	a#(c(c(x1)))	(16)
  a#(c(a(x1)))	→	c#(x1)	(14)
  c#(a(c(x1)))	→	a#(x1)	(17)
  a#(a(a(x1)))	→	c#(x1)	(11)

  1.1.1.1.1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(c#)	=	{ 1, 1 }
  π(a#)	=	{ 1, 1 }
  π(c)	=	{ 1, 1 }
  π(a)	=	{ 1, 1 }

  to remove the pairs:

  c#(a(c(x1)))	→	a#(a(x1))	(18)
  a#(c(a(x1)))	→	c#(x1)	(14)
  c#(a(c(x1)))	→	a#(x1)	(17)
  a#(a(a(x1)))	→	c#(x1)	(11)

  1.1.1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 2 components.

  • The 1^st component contains the pair

    a#(c(a(x1)))

    →

    a#(c(c(x1)))

    (16)

    1.1.1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals

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

    together with the usable rules

    a(a(a(x1)))	→	a(c(c(x1)))	(8)
    a(c(a(x1)))	→	a(c(c(x1)))	(9)
    c(a(c(x1)))	→	c(a(a(x1)))	(10)

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

    a#(c(a(x1)))

    →

    a#(c(c(x1)))

    (16)

    could be deleted.

    1.1.1.1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.

  • The 2^nd component contains the pair

    c#(a(c(x1)))

    →

    c#(a(a(x1)))

    (19)

    1.1.1.1.1.1.1.1.2 Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

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

    together with the usable rules

    a(a(a(x1)))	→	a(c(c(x1)))	(8)
    a(c(a(x1)))	→	a(c(c(x1)))	(9)
    c(a(c(x1)))	→	c(a(a(x1)))	(10)

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

    c#(a(c(x1)))

    →

    c#(a(a(x1)))

    (19)

    could be deleted.

    1.1.1.1.1.1.1.1.2.1 P is empty

    There are no pairs anymore.