Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/213281)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ 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#(2(4(3(5(x1)))))	→	3#(x1)	(103)
  3#(0(1(3(2(x1)))))	→	0#(3(2(x1)))	(108)
  3#(0(4(0(2(x1)))))	→	0#(3(x1))	(117)
  3#(0(4(0(2(x1)))))	→	3#(x1)	(118)
  3#(4(0(1(4(x1)))))	→	3#(4(x1))	(129)
  3#(4(0(1(5(x1)))))	→	3#(x1)	(131)
  3#(4(1(2(4(x1)))))	→	3#(x1)	(136)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  together with the usable rules

  3(2(4(1(2(x1)))))	→	3(1(2(2(5(4(x1))))))	(38)
  3(2(4(1(5(x1)))))	→	3(1(4(5(2(5(x1))))))	(39)
  3(4(0(2(x1))))	→	3(0(4(5(2(x1)))))	(12)
  3(4(0(2(x1))))	→	3(5(0(4(2(x1)))))	(13)
  3(0(1(3(2(x1)))))	→	0(3(1(0(3(2(x1))))))	(31)
  3(0(2(1(4(x1)))))	→	4(0(4(1(3(2(x1))))))	(32)
  3(0(2(1(5(x1)))))	→	5(3(2(0(4(1(x1))))))	(33)
  3(0(4(0(2(x1)))))	→	0(3(4(0(4(2(x1))))))	(34)
  3(0(4(0(2(x1)))))	→	0(4(1(2(0(3(x1))))))	(35)
  3(0(5(1(4(x1)))))	→	3(0(4(1(1(5(x1))))))	(36)
  3(0(5(1(5(x1)))))	→	0(4(1(3(5(5(x1))))))	(37)
  3(4(0(1(2(x1)))))	→	0(4(2(0(3(1(x1))))))	(40)
  3(4(0(1(4(x1)))))	→	0(4(1(5(3(4(x1))))))	(41)
  3(4(0(1(5(x1)))))	→	0(4(1(5(5(3(x1))))))	(42)
  3(4(0(2(4(x1)))))	→	0(3(4(0(4(2(x1))))))	(43)
  3(4(1(2(4(x1)))))	→	0(4(1(2(4(3(x1))))))	(44)
  3(4(1(3(5(x1)))))	→	4(3(0(3(1(5(x1))))))	(45)
  3(4(3(0(2(x1)))))	→	3(3(0(4(1(2(x1))))))	(46)
  3(4(5(0(2(x1)))))	→	0(3(0(4(2(5(x1))))))	(47)
  3(5(0(2(2(x1)))))	→	0(3(2(5(2(5(x1))))))	(48)
  3(5(2(1(4(x1)))))	→	3(5(1(0(4(2(x1))))))	(49)

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

  0#(2(4(3(5(x1)))))

  →

  3#(x1)

  (103)

  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

    3#(4(0(1(4(x1)))))	→	3#(4(x1))	(129)
    3#(4(0(1(5(x1)))))	→	3#(x1)	(131)
    3#(0(4(0(2(x1)))))	→	3#(x1)	(118)
    3#(4(1(2(4(x1)))))	→	3#(x1)	(136)

    1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

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

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

    1.1.1.1.1.1 Size-Change Termination

    Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

    3#(0(4(0(2(x1)))))	→	3#(x1)	(118)
    1	>	1
    3#(4(0(1(4(x1)))))	→	3#(4(x1))	(129)
    1	>	1
    3#(4(0(1(5(x1)))))	→	3#(x1)	(131)
    1	>	1
    3#(4(1(2(4(x1)))))	→	3#(x1)	(136)
    1	>	1

    As there is no critical graph in the transitive closure, there are no infinite chains.