Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex9_Luc06_GM)

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

  mark#(f(X1,X2,X3))

  →

  mark#(X2)

  (8)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a]	=	0
  [b]	=	0
  [a__b]	=	0
  [a__f(x1, x2, x3)]	=	0
  [a__b#]	=	0
  [f(x1, x2, x3)]	=	x2 + 1
  [mark#(x1)]	=	x1 + 0
  [mark(x1)]	=	0
  [a__f#(x1, x2, x3)]	=	0

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

  mark#(f(X1,X2,X3))

  →

  mark#(X2)

  (8)

  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

  a__f#(a,X,X)

  →

  a__f#(X,a__b,b)

  (10)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a]	=	3
  [b]	=	2
  [a__b]	=	1
  [a__f(x1, x2, x3)]	=	0
  [a__b#]	=	0
  [f(x1, x2, x3)]	=	1
  [mark#(x1)]	=	0
  [mark(x1)]	=	0
  [a__f#(x1, x2, x3)]	=	x1 + x3 + 0

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

  a__f#(a,X,X)

  →

  a__f#(X,a__b,b)

  (10)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.