Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/213281)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

There are 101 ruless (increase limit for explicit display).

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))	(149)
  0#(0(1(4(5(x1)))))	→	0#(3(5(x1)))	(108)
  0#(1(2(3(4(x1)))))	→	0#(3(x1))	(102)
  0#(2(1(4(4(x1)))))	→	3#(x1)	(101)
  0#(2(4(3(5(x1)))))	→	3#(x1)	(144)
  0#(1(3(4(x1))))	→	3#(x1)	(115)
  3#(4(0(1(5(x1)))))	→	3#(x1)	(140)
  3#(4(1(2(4(x1)))))	→	3#(x1)	(137)
  3#(0(4(0(2(x1)))))	→	3#(x1)	(84)
  0#(1(2(3(4(x1)))))	→	3#(x1)	(80)
  3#(0(1(3(2(x1)))))	→	0#(3(2(x1)))	(66)
  0#(1(3(4(x1))))	→	3#(x1)	(115)
  3#(0(4(0(2(x1)))))	→	0#(3(x1))	(114)
  0#(2(1(4(x1))))	→	3#(x1)	(56)
  0#(1(3(4(x1))))	→	0#(3(x1))	(50)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [0#(x1)]	=	x1 + 0
  [1(x1)]	=	24391
  [4(x1)]	=	1
  [5(x1)]	=	1
  [3(x1)]	=	1
  [0(x1)]	=	x1 + 0
  [3#(x1)]	=	2
  [2(x1)]	=	24391

  together with the usable rules

  0(1(4(0(2(x1)))))	→	0(4(1(5(0(2(x1))))))	(18)
  0(1(3(4(x1))))	→	0(4(1(3(1(x1)))))	(4)
  0(1(0(2(4(x1)))))	→	2(0(0(4(1(1(x1))))))	(15)
  0(2(1(4(x1))))	→	5(5(0(4(1(2(x1))))))	(8)
  0(1(0(2(x1))))	→	0(0(3(1(2(x1)))))	(1)
  0(1(3(4(x1))))	→	0(4(1(1(3(x1)))))	(3)
  0(1(2(3(4(x1)))))	→	2(0(4(1(0(3(x1))))))	(16)
  0(1(4(3(4(x1)))))	→	3(0(4(1(5(4(x1))))))	(21)
  3(0(5(1(4(x1)))))	→	3(0(4(1(1(5(x1))))))	(36)
  0(2(1(4(5(x1)))))	→	0(4(1(2(5(2(x1))))))	(26)
  0(1(4(1(5(x1)))))	→	2(5(0(4(1(1(x1))))))	(19)
  3(0(2(1(4(x1)))))	→	4(0(4(1(3(2(x1))))))	(32)
  0(1(3(3(4(x1)))))	→	0(0(3(1(3(4(x1))))))	(17)
  0(2(1(5(4(x1)))))	→	5(0(2(0(4(1(x1))))))	(27)
  3(0(4(0(2(x1)))))	→	0(3(4(0(4(2(x1))))))	(34)
  0(1(4(3(5(x1)))))	→	5(4(5(0(3(1(x1))))))	(22)
  0(2(4(1(5(x1)))))	→	5(0(4(1(5(2(x1))))))	(28)
  3(4(1(2(4(x1)))))	→	0(4(1(2(4(3(x1))))))	(44)
  0(2(1(4(x1))))	→	0(4(1(2(3(x1)))))	(5)
  3(0(2(1(5(x1)))))	→	5(3(2(0(4(1(x1))))))	(33)
  0(2(2(4(x1))))	→	0(4(2(2(5(x1)))))	(10)
  3(2(4(1(5(x1)))))	→	3(1(4(5(2(5(x1))))))	(39)
  0(2(1(4(x1))))	→	2(0(4(1(4(x1)))))	(7)
  0(1(4(3(4(x1)))))	→	0(4(0(3(1(4(x1))))))	(20)
  0(2(1(4(4(x1)))))	→	0(4(1(2(4(3(x1))))))	(25)
  3(5(2(1(4(x1)))))	→	3(5(1(0(4(2(x1))))))	(49)
  0(2(5(1(4(x1)))))	→	0(0(5(4(1(2(x1))))))	(30)
  0(0(1(4(5(x1)))))	→	0(4(1(0(3(5(x1))))))	(14)
  3(0(1(3(2(x1)))))	→	0(3(1(0(3(2(x1))))))	(31)
  3(4(0(2(x1))))	→	3(0(4(5(2(x1)))))	(12)
  3(4(1(3(5(x1)))))	→	4(3(0(3(1(5(x1))))))	(45)
  0(1(5(0(2(x1)))))	→	0(0(4(1(2(5(x1))))))	(23)
  0(1(5(1(4(x1)))))	→	4(5(0(3(1(1(x1))))))	(24)
  0(2(2(4(x1))))	→	0(4(2(5(2(x1)))))	(11)
  0(2(1(5(x1))))	→	5(0(4(1(2(x1)))))	(9)
  3(4(0(2(x1))))	→	3(5(0(4(2(x1)))))	(13)
  3(4(0(1(2(x1)))))	→	0(4(2(0(3(1(x1))))))	(40)
  0(2(1(4(x1))))	→	0(4(1(3(2(x1)))))	(6)
  3(2(4(1(2(x1)))))	→	3(1(2(2(5(4(x1))))))	(38)
  3(5(0(2(2(x1)))))	→	0(3(2(5(2(5(x1))))))	(48)
  3(4(5(0(2(x1)))))	→	0(3(0(4(2(5(x1))))))	(47)
  3(0(5(1(5(x1)))))	→	0(4(1(3(5(5(x1))))))	(37)
  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(3(0(2(x1)))))	→	3(3(0(4(1(2(x1))))))	(46)
  3(0(4(0(2(x1)))))	→	0(4(1(2(0(3(x1))))))	(35)
  0(2(4(3(5(x1)))))	→	0(4(5(2(5(3(x1))))))	(29)
  3(4(0(2(4(x1)))))	→	0(3(4(0(4(2(x1))))))	(43)
  0(1(3(4(x1))))	→	0(4(1(0(3(x1)))))	(2)

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

  0#(0(1(4(5(x1)))))	→	0#(3(5(x1)))	(108)
  0#(1(2(3(4(x1)))))	→	0#(3(x1))	(102)
  0#(2(1(4(4(x1)))))	→	3#(x1)	(101)
  0#(2(4(3(5(x1)))))	→	3#(x1)	(144)
  0#(1(3(4(x1))))	→	3#(x1)	(115)
  0#(1(2(3(4(x1)))))	→	3#(x1)	(80)
  3#(0(1(3(2(x1)))))	→	0#(3(2(x1)))	(66)
  0#(1(3(4(x1))))	→	3#(x1)	(115)
  3#(0(4(0(2(x1)))))	→	0#(3(x1))	(114)
  0#(2(1(4(x1))))	→	3#(x1)	(56)
  0#(1(3(4(x1))))	→	0#(3(x1))	(50)

  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(1(2(4(x1)))))	→	3#(x1)	(137)
    3#(4(0(1(4(x1)))))	→	3#(4(x1))	(149)
    3#(4(0(1(5(x1)))))	→	3#(x1)	(140)
    3#(0(4(0(2(x1)))))	→	3#(x1)	(84)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [0#(x1)]	=	0
    [1(x1)]	=	x1 + 14132
    [4(x1)]	=	x1 + 27570
    [5(x1)]	=	x1 + 7066
    [3(x1)]	=	7067
    [0(x1)]	=	x1 + 0
    [3#(x1)]	=	x1 + 2
    [2(x1)]	=	x1 + 0

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

    3#(4(1(2(4(x1)))))	→	3#(x1)	(137)
    3#(4(0(1(4(x1)))))	→	3#(4(x1))	(149)
    3#(4(0(1(5(x1)))))	→	3#(x1)	(140)
    3#(0(4(0(2(x1)))))	→	3#(x1)	(84)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.