Certification Problem

Input (TPDB SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-559)

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

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

  [c(x1)]	=	-∞ 0 -∞ -∞ 4 0 0 8 4 · x1 + 3 1 10
  [b(x1)]	=	-∞ -∞ 0 0 -∞ -∞ -∞ 0 -∞ · x1 + 1 -∞ -∞
  [a(x1)]	=	4 0 8 -∞ -∞ 0 0 -∞ 4 · x1 + 9 2 1
  [b#(x1)]	=	-∞ -∞ 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 8 -∞ -∞
  [a#(x1)]	=	-∞ -∞ 9 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 1 -∞ -∞

  together with the usable rules

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

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

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

  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

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

  together with the usable rules

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

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

  b#(c(x1))

  →

  a#(b(x1))

  (5)

  and no rules could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.