Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z26)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  f#(a(x),a(y))	→	a#(f(x,y))	(11)
  a#(a(f(x,y)))	→	f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))	(4)
  f#(a(x),a(y))	→	f#(x,y)	(12)
  f#(b(x),b(y))	→	f#(x,y)	(13)
  a#(a(f(x,y)))	→	a#(x)	(7)
  a#(a(f(x,y)))	→	a#(y)	(10)

  1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [a(x1)]	=	1 · x1
  [f(x1, x2)]	=	1 + 1 · x1 + 1 · x2
  [b(x1)]	=	1 · x1
  [f#(x1, x2)]	=	2 + 2 · x1 + 2 · x2
  [a#(x1)]	=	2 · x1

  the pairs

  a#(a(f(x,y)))	→	a#(x)	(7)
  a#(a(f(x,y)))	→	a#(y)	(10)

  and no rules could be deleted.

  1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

  [f#(x1, x2)]	=	0 0 + 1 0 0 0 · x1 + 0 0 0 0 · x2
  [a(x1)]	=	0 1 + 1 1 0 0 · x1
  [a#(x1)]	=	0 0 + 1 1 0 0 · x1
  [f(x1, x2)]	=	0 0 + 1 0 0 1 · x1 + 0 0 0 0 · x2
  [b(x1)]	=	0 0 + 1 0 0 0 · x1

  the pair

  a#(a(f(x,y)))

  →

  f#(a(b(a(b(a(x))))),a(b(a(b(a(y))))))

  (4)

  could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    f#(b(x),b(y))	→	f#(x,y)	(13)
    f#(a(x),a(y))	→	f#(x,y)	(12)

    1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

    [b(x1)]	=	1 · x1
    [a(x1)]	=	1 · x1
    [f#(x1, x2)]	=	1 · x1 + 1 · x2

    having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

    1.1.1.1.1.1.1 Size-Change Termination

    Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

    f#(b(x),b(y))	→	f#(x,y)	(13)
    1	>	1
    2	>	2
    f#(a(x),a(y))	→	f#(x,y)	(12)
    1	>	1
    2	>	2

    As there is no critical graph in the transitive closure, there are no infinite chains.