Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/214091)

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 2 components.

• The 1^st component contains the pair

  2#(4(0(5(1(x1)))))	→	2#(x1)	(70)
  2#(5(2(0(1(x1)))))	→	2#(x1)	(56)
  2#(5(1(x1)))	→	2#(x1)	(59)

  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

  [5(x1)]	=	x1 + 1
  [4(x1)]	=	x1 + 1
  [2(x1)]	=	x1 + 1
  [1(x1)]	=	x1 + 1
  [0(x1)]	=	x1 + 1
  [2#(x1)]	=	x1 + 0

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

  2#(4(0(5(1(x1)))))	→	2#(x1)	(70)
  2#(5(2(0(1(x1)))))	→	2#(x1)	(56)
  2#(5(1(x1)))	→	2#(x1)	(59)

  and no rules 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

  0#(1(2(x1)))	→	0#(x1)	(137)
  0#(5(5(2(x1))))	→	0#(x1)	(111)
  0#(2(4(3(1(x1)))))	→	0#(x1)	(123)

  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

  [5(x1)]	=	x1 + 1
  [4(x1)]	=	x1 + 1
  [3(x1)]	=	x1 + 1
  [2(x1)]	=	x1 + 1
  [1(x1)]	=	x1 + 1
  [0#(x1)]	=	x1 + 0

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

  0#(1(2(x1)))	→	0#(x1)	(137)
  0#(5(5(2(x1))))	→	0#(x1)	(111)
  0#(2(4(3(1(x1)))))	→	0#(x1)	(123)

  and no rules could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.