Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_2_Luc02c_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 1 component.

• The 1^st component contains the pair

  mark#(2nd(X))	→	a__2nd#(mark(X))	(16)
  mark#(from(X))	→	mark#(X)	(15)
  a__from#(X)	→	mark#(X)	(14)
  mark#(cons(X1,X2))	→	mark#(X1)	(13)
  a__2nd#(cons(X,cons(Y,Z)))	→	mark#(Y)	(10)
  mark#(2nd(X))	→	mark#(X)	(12)
  mark#(from(X))	→	a__from#(mark(X))	(11)
  mark#(s(X))	→	mark#(X)	(9)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	x1 + 0
  [a__from#(x1)]	=	x1 + 1
  [a__from(x1)]	=	x1 + 11800
  [2nd(x1)]	=	x1 + 3
  [a__2nd(x1)]	=	x1 + 3
  [a__2nd#(x1)]	=	x1 + 1
  [mark#(x1)]	=	x1 + 0
  [from(x1)]	=	x1 + 11800
  [mark(x1)]	=	x1 + 1
  [cons(x1, x2)]	=	max(x1 + 8366, x2 + 0, 0)

  together with the usable rules

  mark(from(X))	→	a__from(mark(X))	(4)
  a__from(X)	→	from(X)	(8)
  a__2nd(cons(X,cons(Y,Z)))	→	mark(Y)	(1)
  mark(2nd(X))	→	a__2nd(mark(X))	(3)
  mark(cons(X1,X2))	→	cons(mark(X1),X2)	(5)
  a__2nd(X)	→	2nd(X)	(7)
  mark(s(X))	→	s(mark(X))	(6)
  a__from(X)	→	cons(mark(X),from(s(X)))	(2)

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

  mark#(2nd(X))	→	a__2nd#(mark(X))	(16)
  mark#(from(X))	→	mark#(X)	(15)
  a__from#(X)	→	mark#(X)	(14)
  mark#(cons(X1,X2))	→	mark#(X1)	(13)
  a__2nd#(cons(X,cons(Y,Z)))	→	mark#(Y)	(10)
  mark#(2nd(X))	→	mark#(X)	(12)
  mark#(from(X))	→	a__from#(mark(X))	(11)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    mark#(s(X))

    →

    mark#(X)

    (9)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 1
    [a__from#(x1)]	=	2
    [a__from(x1)]	=	7
    [2nd(x1)]	=	1
    [a__2nd(x1)]	=	x1 + 0
    [a__2nd#(x1)]	=	0
    [mark#(x1)]	=	x1 + 2
    [from(x1)]	=	1
    [mark(x1)]	=	x1 + 5
    [cons(x1, x2)]	=	x1 + x2 + 2

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

    mark#(s(X))

    →

    mark#(X)

    (9)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.