Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z121)

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

  d#(b(x1))	→	a#(b(x1))	(25)
  b#(d(x1))	→	d#(b(x1))	(24)
  f#(f(x1))	→	b#(b(b(x1)))	(14)
  f#(f(x1))	→	b#(b(x1))	(23)
  b#(b(x1))	→	a#(c(x1))	(13)
  c#(c(x1))	→	d#(d(x1))	(22)
  a#(f(x1))	→	f#(a(a(x1)))	(21)
  a#(f(x1))	→	a#(x1)	(20)
  f#(f(x1))	→	b#(x1)	(19)
  b#(b(x1))	→	c#(a(c(x1)))	(12)
  c#(c(x1))	→	d#(x1)	(11)
  b#(b(x1))	→	c#(x1)	(10)
  d#(b(x1))	→	d#(a(b(x1)))	(18)
  c#(d(d(x1)))	→	f#(x1)	(9)
  b#(d(x1))	→	b#(x1)	(17)
  b#(b(x1))	→	c#(c(a(c(x1))))	(16)
  c#(c(x1))	→	d#(d(d(x1)))	(15)
  a#(f(x1))	→	a#(a(x1))	(8)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1)]	=	x1 + 0
  [d(x1)]	=	x1 + 12000
  [b(x1)]	=	x1 + 27000
  [c(x1)]	=	x1 + 18000
  [f(x1)]	=	x1 + 42000
  [d#(x1)]	=	x1 + 1
  [f#(x1)]	=	x1 + 27004
  [c#(x1)]	=	x1 + 6002
  [a#(x1)]	=	x1 + 0
  [b#(x1)]	=	x1 + 15003

  together with the usable rules

  d(b(x1))	→	d(a(b(x1)))	(4)
  f(f(x1))	→	b(b(b(x1)))	(1)
  b(b(x1))	→	c(c(a(c(x1))))	(3)
  c(c(x1))	→	d(d(d(x1)))	(5)
  c(d(d(x1)))	→	f(x1)	(7)
  b(d(x1))	→	d(b(x1))	(6)
  a(f(x1))	→	f(a(a(x1)))	(2)

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

  d#(b(x1))	→	a#(b(x1))	(25)
  b#(d(x1))	→	d#(b(x1))	(24)
  f#(f(x1))	→	b#(b(b(x1)))	(14)
  f#(f(x1))	→	b#(b(x1))	(23)
  b#(b(x1))	→	a#(c(x1))	(13)
  c#(c(x1))	→	d#(d(x1))	(22)
  a#(f(x1))	→	f#(a(a(x1)))	(21)
  a#(f(x1))	→	a#(x1)	(20)
  f#(f(x1))	→	b#(x1)	(19)
  b#(b(x1))	→	c#(a(c(x1)))	(12)
  c#(c(x1))	→	d#(x1)	(11)
  b#(b(x1))	→	c#(x1)	(10)
  c#(d(d(x1)))	→	f#(x1)	(9)
  b#(d(x1))	→	b#(x1)	(17)
  b#(b(x1))	→	c#(c(a(c(x1))))	(16)
  c#(c(x1))	→	d#(d(d(x1)))	(15)
  a#(f(x1))	→	a#(a(x1))	(8)

  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

    d#(b(x1))

    →

    d#(a(b(x1)))

    (18)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the argument filter

    π(a#)

    =

    1

    in combination with the following Weighted Path Order with the following precedence and status

    prec(a)	=	0		status(a)	=	[]		list-extension(a)	=	Lex
    prec(d)	=	0		status(d)	=	[]		list-extension(d)	=	Lex
    prec(b)	=	1		status(b)	=	[]		list-extension(b)	=	Lex
    prec(c)	=	0		status(c)	=	[1]		list-extension(c)	=	Lex
    prec(f)	=	0		status(f)	=	[]		list-extension(f)	=	Lex
    prec(d#)	=	0		status(d#)	=	[1]		list-extension(d#)	=	Lex
    prec(f#)	=	0		status(f#)	=	[]		list-extension(f#)	=	Lex
    prec(c#)	=	0		status(c#)	=	[]		list-extension(c#)	=	Lex
    prec(b#)	=	0		status(b#)	=	[]		list-extension(b#)	=	Lex

    and the following Max-polynomial interpretation

    [a(x1)]	=	x1 + 0
    [d(x1)]	=	x1 + 193733
    [b(x1)]	=	x1 + 452043
    [c(x1)]	=	x1 + 290600
    [f(x1)]	=	x1 + 678065
    [d#(x1)]	=	x1 + 1
    [f#(x1)]	=	1
    [c#(x1)]	=	1
    [b#(x1)]	=	1

    together with the usable rules

    d(b(x1))	→	d(a(b(x1)))	(4)
    f(f(x1))	→	b(b(b(x1)))	(1)
    b(b(x1))	→	c(c(a(c(x1))))	(3)
    c(c(x1))	→	d(d(d(x1)))	(5)
    c(d(d(x1)))	→	f(x1)	(7)
    b(d(x1))	→	d(b(x1))	(6)
    a(f(x1))	→	f(a(a(x1)))	(2)

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

    d#(b(x1))

    →

    d#(a(b(x1)))

    (18)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.