Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-552)

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 2 components.

• The 1^st component contains the pair

  b#(b(a(a(x1))))	→	b#(a(a(a(x1))))	(10)
  b#(b(a(a(x1))))	→	b#(a(b(b(x1))))	(9)
  b#(b(a(a(x1))))	→	b#(x1)	(5)
  b#(b(a(a(x1))))	→	b#(b(x1))	(8)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 28958
  [b(x1)]	=	x1 + 28958
  [a#(x1)]	=	0
  [b#(x1)]	=	x1 + 0

  together with the usable rules

  a(b(b(a(x1))))	→	a(a(a(a(x1))))	(1)
  b(b(a(a(x1))))	→	b(a(b(b(x1))))	(3)
  b(b(a(a(x1))))	→	b(a(a(a(x1))))	(2)

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

  b#(b(a(a(x1))))	→	b#(x1)	(5)
  b#(b(a(a(x1))))	→	b#(b(x1))	(8)

  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

    b#(b(a(a(x1))))	→	b#(a(b(b(x1))))	(9)
    b#(b(a(a(x1))))	→	b#(a(a(a(x1))))	(10)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a(x1)]	=	20166
    [b(x1)]	=	x1 + 1
    [a#(x1)]	=	0
    [b#(x1)]	=	x1 + 0

    together with the usable rules

    a(b(b(a(x1))))	→	a(a(a(a(x1))))	(1)
    b(b(a(a(x1))))	→	b(a(b(b(x1))))	(3)
    b(b(a(a(x1))))	→	b(a(a(a(x1))))	(2)

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

    b#(b(a(a(x1))))	→	b#(a(b(b(x1))))	(9)
    b#(b(a(a(x1))))	→	b#(a(a(a(x1))))	(10)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  a#(b(b(a(x1))))	→	a#(a(a(x1)))	(11)
  a#(b(b(a(x1))))	→	a#(a(x1))	(6)
  a#(b(b(a(x1))))	→	a#(a(a(a(x1))))	(7)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	1
  [b(x1)]	=	x1 + 42736
  [a#(x1)]	=	x1 + 0
  [b#(x1)]	=	x1 + 0

  together with the usable rules

  a(b(b(a(x1))))	→	a(a(a(a(x1))))	(1)
  b(b(a(a(x1))))	→	b(a(b(b(x1))))	(3)
  b(b(a(a(x1))))	→	b(a(a(a(x1))))	(2)

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

  a#(b(b(a(x1))))	→	a#(a(a(x1)))	(11)
  a#(b(b(a(x1))))	→	a#(a(x1))	(6)
  a#(b(b(a(x1))))	→	a#(a(a(a(x1))))	(7)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.