Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/160263)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  5#(0(1(2(2(x1)))))	→	5#(2(x1))	(130)
  5#(2(4(0(5(x1)))))	→	5#(5(5(x1)))	(138)
  5#(2(4(0(5(x1)))))	→	5#(5(x1))	(137)
  5#(0(1(2(2(x1)))))	→	0#(5(2(x1)))	(131)
  0#(5(1(2(2(x1)))))	→	5#(2(x1))	(120)
  0#(1(4(5(5(x1)))))	→	0#(1(5(x1)))	(115)
  0#(1(4(5(2(x1)))))	→	0#(5(x1))	(113)
  0#(1(4(5(2(x1)))))	→	5#(x1)	(112)
  5#(1(2(2(x1))))	→	5#(x1)	(87)
  0#(1(2(2(2(x1)))))	→	5#(2(x1))	(106)
  0#(1(4(5(x1))))	→	5#(5(0(x1)))	(85)
  0#(1(4(5(x1))))	→	5#(0(x1))	(83)
  0#(1(4(5(x1))))	→	0#(x1)	(82)
  0#(1(2(2(x1))))	→	0#(5(x1))	(76)
  0#(1(2(2(x1))))	→	5#(x1)	(75)
  0#(1(2(2(x1))))	→	0#(x1)	(68)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the

  prec(5#)	=	0		stat(5#)	=	lex
  prec(0#)	=	0		stat(0#)	=	lex
  prec(5)	=	0		stat(5)	=	lex
  prec(4)	=	0		stat(4)	=	lex
  prec(3)	=	0		stat(3)	=	lex
  prec(0)	=	0		stat(0)	=	lex
  prec(1)	=	0		stat(1)	=	lex
  prec(2)	=	0		stat(2)	=	lex

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

  together with the usable rules

  0(1(2(2(x1))))	→	0(1(0(2(2(x1)))))	(1)
  0(1(2(2(x1))))	→	0(1(2(3(2(x1)))))	(2)
  0(1(2(2(x1))))	→	0(2(2(1(3(x1)))))	(3)
  0(1(2(2(x1))))	→	1(0(3(2(2(x1)))))	(4)
  0(1(2(2(x1))))	→	1(2(0(3(2(x1)))))	(5)
  0(1(2(2(x1))))	→	1(3(0(2(2(x1)))))	(6)
  0(1(2(2(x1))))	→	1(3(2(0(2(x1)))))	(7)
  0(1(2(2(x1))))	→	0(1(0(4(2(2(x1))))))	(8)
  0(1(2(2(x1))))	→	0(2(1(3(2(3(x1))))))	(9)
  0(1(2(2(x1))))	→	1(2(1(0(4(2(x1))))))	(10)
  0(1(2(2(x1))))	→	1(5(0(4(2(2(x1))))))	(11)
  0(1(2(2(x1))))	→	2(0(3(1(3(2(x1))))))	(12)
  0(1(2(2(x1))))	→	2(1(1(0(4(2(x1))))))	(13)
  0(1(2(2(x1))))	→	2(1(3(0(2(0(x1))))))	(14)
  0(1(2(2(x1))))	→	2(1(3(3(2(0(x1))))))	(15)
  0(1(2(2(x1))))	→	2(1(5(3(0(2(x1))))))	(16)
  0(1(2(2(x1))))	→	2(2(1(3(0(5(x1))))))	(17)
  0(1(2(2(x1))))	→	2(4(1(3(2(0(x1))))))	(18)
  0(1(4(5(x1))))	→	1(5(0(4(1(x1)))))	(19)
  0(1(4(5(x1))))	→	5(0(4(1(5(x1)))))	(20)
  0(1(4(5(x1))))	→	5(4(1(5(0(x1)))))	(21)
  0(1(4(5(x1))))	→	1(1(5(0(4(1(x1))))))	(22)
  0(1(4(5(x1))))	→	5(4(1(5(5(0(x1))))))	(23)
  5(1(2(2(x1))))	→	1(0(2(2(5(x1)))))	(24)
  5(1(2(2(x1))))	→	1(3(5(2(2(x1)))))	(25)
  5(1(2(2(x1))))	→	1(5(2(3(2(x1)))))	(26)
  5(1(2(2(x1))))	→	1(5(0(2(2(3(x1))))))	(27)
  5(1(2(2(x1))))	→	2(1(0(3(2(5(x1))))))	(28)
  5(1(2(2(x1))))	→	3(1(3(5(2(2(x1))))))	(29)
  5(1(2(2(x1))))	→	4(1(3(2(2(5(x1))))))	(30)
  5(1(2(2(x1))))	→	5(1(0(4(2(2(x1))))))	(31)
  5(1(2(2(x1))))	→	5(1(2(0(4(2(x1))))))	(32)
  0(1(1(4(5(x1)))))	→	3(1(0(4(1(5(x1))))))	(33)
  0(1(2(2(2(x1)))))	→	1(0(2(2(5(2(x1))))))	(34)
  0(1(2(2(5(x1)))))	→	1(5(0(4(2(2(x1))))))	(35)
  0(1(2(4(5(x1)))))	→	2(5(1(0(4(5(x1))))))	(36)
  0(1(4(5(2(x1)))))	→	1(0(4(2(0(5(x1))))))	(37)
  0(1(4(5(5(x1)))))	→	5(0(4(0(1(5(x1))))))	(38)
  0(1(5(4(5(x1)))))	→	1(5(0(4(1(5(x1))))))	(39)
  0(5(1(2(2(x1)))))	→	0(1(3(2(5(2(x1))))))	(40)
  3(3(1(2(2(x1)))))	→	1(3(2(0(3(2(x1))))))	(41)
  3(4(4(0(5(x1)))))	→	3(5(4(5(0(4(x1))))))	(42)
  5(0(1(2(2(x1)))))	→	1(3(2(0(5(2(x1))))))	(43)
  5(1(2(2(5(x1)))))	→	1(5(2(3(2(5(x1))))))	(44)
  5(2(1(2(2(x1)))))	→	2(1(3(5(2(2(x1))))))	(45)
  5(2(4(0(5(x1)))))	→	0(4(2(5(5(5(x1))))))	(46)
  5(2(4(0(5(x1)))))	→	0(4(5(4(2(5(x1))))))	(47)

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

  5#(0(1(2(2(x1)))))	→	5#(2(x1))	(130)
  5#(2(4(0(5(x1)))))	→	5#(5(5(x1)))	(138)
  5#(2(4(0(5(x1)))))	→	5#(5(x1))	(137)
  5#(0(1(2(2(x1)))))	→	0#(5(2(x1)))	(131)
  0#(5(1(2(2(x1)))))	→	5#(2(x1))	(120)
  0#(1(4(5(2(x1)))))	→	0#(5(x1))	(113)
  0#(1(4(5(2(x1)))))	→	5#(x1)	(112)
  5#(1(2(2(x1))))	→	5#(x1)	(87)
  0#(1(2(2(2(x1)))))	→	5#(2(x1))	(106)
  0#(1(2(2(x1))))	→	0#(5(x1))	(76)
  0#(1(2(2(x1))))	→	5#(x1)	(75)
  0#(1(2(2(x1))))	→	0#(x1)	(68)

  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

    0#(1(4(5(5(x1)))))	→	0#(1(5(x1)))	(115)
    0#(1(4(5(x1))))	→	0#(x1)	(82)

    1.1.1.1.1 Subterm Criterion Processor

    We use the projection to multisets

    π(0#)	=	{ 1 }
    π(4)	=	{ 1 }
    π(1)	=	{ 1 }

    to remove the pairs:

    0#(1(4(5(5(x1)))))	→	0#(1(5(x1)))	(115)
    0#(1(4(5(x1))))	→	0#(x1)	(82)

    1.1.1.1.1.1 P is empty

    There are no pairs anymore.