Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove00)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  c#(s(x1))	→	a#(p(s(p(s(x1)))))	(6)
  a#(s(x1))	→	b#(p(p(s(s(x1)))))	(16)
  b#(s(x1))	→	c#(p(s(p(s(x1)))))	(11)

  1.1.1.1 Reduction Pair Processor with Usable Rules

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

  [s(x1)]	=	0 2 0 2 · x1 + -∞ -∞
  [p(x1)]	=	0 0 -2 -2 · x1 + -∞ -∞
  [c#(x1)]	=	3 3 3 3 · x1 + -∞ -∞
  [b#(x1)]	=	3 5 3 5 · x1 + -∞ -∞
  [a#(x1)]	=	3 5 3 5 · x1 + -∞ -∞

  together with the usable rules

  p(p(s(x1)))	→	p(x1)	(4)
  p(s(x1))	→	x1	(5)

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

  b#(s(x1))

  →

  c#(p(s(p(s(x1)))))

  (11)

  could be deleted.

  1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

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

  [s(x1)]	=	x1 + 1
  [p(x1)]	=	x1 + 1
  [c#(x1)]	=	x1 + 8
  [b#(x1)]	=	x1 + 0
  [a#(x1)]	=	x1 + 4

  together with the usable rules

  p(p(s(x1)))	→	p(x1)	(4)
  p(s(x1))	→	x1	(5)

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

  c#(s(x1))	→	a#(p(s(p(s(x1)))))	(6)
  a#(s(x1))	→	b#(p(p(s(s(x1)))))	(16)

  and the rules

  p(p(s(x1)))	→	p(x1)	(4)
  p(s(x1))	→	x1	(5)

  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

  p#(p(s(x1)))

  →

  p#(x1)

  (20)

  1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules

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

  [s(x1)]	=	x1 + 1
  [p(x1)]	=	x1 + 1
  [p#(x1)]	=	x1 + 0

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

  p#(p(s(x1)))

  →

  p#(x1)

  (20)

  and no rules could be deleted.

  1.1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.