Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/26069)

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

  0#(1(2(1(x1))))	→	0#(1(2(x1)))	(5)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(x1))))))	(6)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(0(1(2(x1)))))))))	(7)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(8)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))	(9)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [2(x1)]	=	0 0 1 0 · x1 + 0 0
  [1(x1)]	=	1 0 0 1 · x1 + 1 0
  [0(x1)]	=	0 1 0 1 · x1 + 0 1
  [0#(x1)]	=	0 2 0 0 · x1 + 1 0

  together with the usable rules

  0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(x1))))))))))	(1)
  0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(x1)))))))))))))	(2)
  0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))	(3)
  0(1(2(1(x1))))	→	1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))))	(4)

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

  0#(1(2(1(x1))))	→	0#(1(2(x1)))	(5)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(x1))))))	(6)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(0(1(2(x1)))))))))	(7)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))	(8)
  0#(1(2(1(x1))))	→	0#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))	(9)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.