Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/cime2)

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

  msubst#(msubst(a,s),t)	→	msubst#(a,circ(s,t))	(20)
  circ#(cons(a,s),t)	→	circ#(s,t)	(19)
  circ#(cons(lift,s),cons(a,t))	→	circ#(s,t)	(18)
  circ#(cons(lift,s),circ(cons(lift,t),u))	→	circ#(s,t)	(17)
  circ#(circ(s,t),u)	→	circ#(s,circ(t,u))	(13)
  circ#(cons(lift,s),cons(lift,t))	→	circ#(s,t)	(12)
  msubst#(msubst(a,s),t)	→	circ#(s,t)	(16)
  circ#(circ(s,t),u)	→	circ#(t,u)	(15)
  circ#(cons(a,s),t)	→	msubst#(a,t)	(14)
  circ#(cons(lift,s),circ(cons(lift,t),u))	→	circ#(cons(lift,circ(s,t)),u)	(11)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [msubst#(x1, x2)]	=	x1 + x2 + 0
  [lift]	=	0
  [msubst(x1, x2)]	=	x1 + x2 + 8856
  [subst(x1, x2)]	=	0
  [circ#(x1, x2)]	=	x1 + x2 + 8855
  [subst#(x1, x2)]	=	0
  [cons(x1, x2)]	=	max(x1 + 14235, x2 + 28471, 0)
  [id]	=	0
  [circ(x1, x2)]	=	x1 + x2 + 8856

  together with the usable rules

  circ(circ(s,t),u)	→	circ(s,circ(t,u))	(4)
  circ(cons(a,s),t)	→	cons(msubst(a,t),circ(s,t))	(1)
  circ(cons(lift,s),cons(lift,t))	→	cons(lift,circ(s,t))	(3)
  circ(s,id)	→	s	(5)
  msubst(msubst(a,s),t)	→	msubst(a,circ(s,t))	(10)
  circ(cons(lift,s),circ(cons(lift,t),u))	→	circ(cons(lift,circ(s,t)),u)	(7)
  msubst(a,id)	→	a	(9)
  circ(id,s)	→	s	(6)
  circ(cons(lift,s),cons(a,t))	→	cons(a,circ(s,t))	(2)

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

  circ#(cons(a,s),t)	→	circ#(s,t)	(19)
  circ#(cons(lift,s),cons(a,t))	→	circ#(s,t)	(18)
  circ#(cons(lift,s),circ(cons(lift,t),u))	→	circ#(s,t)	(17)
  circ#(cons(lift,s),cons(lift,t))	→	circ#(s,t)	(12)
  msubst#(msubst(a,s),t)	→	circ#(s,t)	(16)
  circ#(circ(s,t),u)	→	circ#(t,u)	(15)
  circ#(cons(a,s),t)	→	msubst#(a,t)	(14)
  circ#(cons(lift,s),circ(cons(lift,t),u))	→	circ#(cons(lift,circ(s,t)),u)	(11)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 2 components.

  • The 1^st component contains the pair

    circ#(circ(s,t),u)

    →

    circ#(s,circ(t,u))

    (13)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [msubst#(x1, x2)]	=	0
    [lift]	=	1
    [msubst(x1, x2)]	=	x1 + 2
    [subst(x1, x2)]	=	0
    [circ#(x1, x2)]	=	x1 + 0
    [subst#(x1, x2)]	=	0
    [cons(x1, x2)]	=	x1 + 14681
    [id]	=	1
    [circ(x1, x2)]	=	x1 + 1

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

    circ#(circ(s,t),u)

    →

    circ#(s,circ(t,u))

    (13)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.

  • The 2^nd component contains the pair

    msubst#(msubst(a,s),t)

    →

    msubst#(a,circ(s,t))

    (20)

    1.1.1.1.2 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [msubst#(x1, x2)]	=	x1 + 0
    [lift]	=	1
    [msubst(x1, x2)]	=	x1 + 44365
    [subst(x1, x2)]	=	0
    [circ#(x1, x2)]	=	x1 + 0
    [subst#(x1, x2)]	=	0
    [cons(x1, x2)]	=	x1 + 14681
    [id]	=	1
    [circ(x1, x2)]	=	x1 + 44364

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

    msubst#(msubst(a,s),t)

    →

    msubst#(a,circ(s,t))

    (20)

    could be deleted.

    1.1.1.1.2.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.