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 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

  a#(s(x1))	→	b#(p(p(s(s(x1)))))	(11)
  c#(s(x1))	→	a#(p(s(p(s(x1)))))	(18)
  b#(s(x1))	→	c#(p(s(p(s(x1)))))	(16)

  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)]	=	0 0
  [s(x1)]	=	1 1 1 0 · x1 + 17 1
  [b(x1)]	=	0 0
  [c(x1)]	=	0 0
  [p#(x1)]	=	0 0
  [p(x1)]	=	0 1 1 0 · x1 + 0 10
  [c#(x1)]	=	1 0 0 0 · x1 + 0 6
  [a#(x1)]	=	1 0 0 0 · x1 + 14 6
  [b#(x1)]	=	0 1 0 0 · x1 + 2 6

  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

  a#(s(x1))	→	b#(p(p(s(s(x1)))))	(11)
  c#(s(x1))	→	a#(p(s(p(s(x1)))))	(18)
  b#(s(x1))	→	c#(p(s(p(s(x1)))))	(16)

  could be deleted.

  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)

  (19)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

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

  (19)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.