Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/3)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ 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

  a#(a(x1))	→	b#(a(x1))	(6)
  a#(a(x1))	→	a#(b(a(x1)))	(4)
  b#(a(b(x1)))	→	a#(x1)	(3)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 21240
  [b(x1)]	=	x1 + 0
  [c(x1)]	=	0
  [a#(x1)]	=	x1 + 21239
  [b#(x1)]	=	x1 + 0

  together with the usable rules

  a(a(x1))	→	a(b(a(x1)))	(1)
  b(a(b(x1)))	→	a(c(a(x1)))	(2)

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

  a#(a(x1))	→	b#(a(x1))	(6)
  b#(a(b(x1)))	→	a#(x1)	(3)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    a#(a(x1))

    →

    a#(b(a(x1)))

    (4)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

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

    [a(x1)]	=	1 0 1 0 · x1 + 6 1
    [b(x1)]	=	0 1 0 1 · x1 + 3 1
    [c(x1)]	=	1 1
    [a#(x1)]	=	1 1 0 0 · x1 + 0 0
    [b#(x1)]	=	0 0

    together with the usable rules

    a(a(x1))	→	a(b(a(x1)))	(1)
    b(a(b(x1)))	→	a(c(a(x1)))	(2)

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

    a#(a(x1))

    →

    a#(b(a(x1)))

    (4)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.