Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/18)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by matchbox @ 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

  b#(b(x1))	→	a#(x1)	(4)
  a#(b(a(x1)))	→	b#(b(a(x1)))	(9)
  b#(b(x1))	→	a#(a(x1))	(5)
  b#(b(x1))	→	a#(a(a(x1)))	(6)
  b#(b(b(x1)))	→	b#(a(x1))	(7)
  b#(b(b(x1)))	→	a#(x1)	(8)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [b(x1)]	=	0 2 0 0 · x1 + -∞ -∞
  [a(x1)]	=	0 2 -2 0 · x1 + -∞ -∞
  [b#(x1)]	=	7 8 7 8 · x1 + -∞ -∞
  [a#(x1)]	=	7 9 7 9 · x1 + -∞ -∞

  together with the usable rules

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

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

  a#(b(a(x1)))	→	b#(b(a(x1)))	(9)
  b#(b(b(x1)))	→	b#(a(x1))	(7)
  b#(b(b(x1)))	→	a#(x1)	(8)

  could be deleted.

  1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1

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

  together with the usable rules

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

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

  b#(b(x1))	→	a#(x1)	(4)
  b#(b(x1))	→	a#(a(x1))	(5)
  b#(b(x1))	→	a#(a(a(x1)))	(6)

  and no rules could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.