Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/pair3rotate)

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

  p#(a(x0),p(b(x1),p(a(x2),x3)))	→	p#(x2,p(a(a(x0)),p(b(x1),x3)))	(4)
  p#(a(x0),p(b(x1),p(a(x2),x3)))	→	p#(a(a(x0)),p(b(x1),x3))	(2)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 30613
  [b(x1)]	=	42736
  [p#(x1, x2)]	=	x1 + x2 + 0
  [p(x1, x2)]	=	x1 + x2 + 1

  together with the usable rule

  p(a(x0),p(b(x1),p(a(x2),x3)))

  →

  p(x2,p(a(a(x0)),p(b(x1),x3)))

  (1)

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

  p#(a(x0),p(b(x1),p(a(x2),x3)))

  →

  p#(a(a(x0)),p(b(x1),x3))

  (2)

  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

    p#(a(x0),p(b(x1),p(a(x2),x3)))

    →

    p#(x2,p(a(a(x0)),p(b(x1),x3)))

    (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 1 0 0 · x1 + 0 0
    [b(x1)]	=	1 1 1 1 · x1 + 0 1
    [p#(x1, x2)]	=	1 0 1 1 · x1 + 0 1 1 0 · x2 + 0 0
    [p(x1, x2)]	=	1 0 1 1 · x1 + 1 0 1 0 · x2 + 0 45001

    together with the usable rule

    p(a(x0),p(b(x1),p(a(x2),x3)))

    →

    p(x2,p(a(a(x0)),p(b(x1),x3)))

    (1)

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

    p#(a(x0),p(b(x1),p(a(x2),x3)))

    →

    p#(x2,p(a(a(x0)),p(b(x1),x3)))

    (4)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.