Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.33)

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 2 components.

• The 1^st component contains the pair

  weight#(cons(n,cons(m,x)))

  →

  weight#(sum(cons(n,cons(m,x)),cons(0,x)))

  (7)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	1
  [weight#(x1)]	=	x1 + 0
  [sum(x1, x2)]	=	x2 + 0
  [0]	=	1
  [nil]	=	1
  [weight(x1)]	=	0
  [cons(x1, x2)]	=	x2 + 1
  [sum#(x1, x2)]	=	0

  together with the usable rules

  sum(cons(s(n),x),cons(m,y))	→	sum(cons(n,x),cons(s(m),y))	(1)
  sum(nil,y)	→	y	(3)
  sum(cons(0,x),y)	→	sum(x,y)	(2)

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

  weight#(cons(n,cons(m,x)))

  →

  weight#(sum(cons(n,cons(m,x)),cons(0,x)))

  (7)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  sum#(cons(0,x),y)	→	sum#(x,y)	(8)
  sum#(cons(s(n),x),cons(m,y))	→	sum#(cons(n,x),cons(s(m),y))	(6)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [s(x1)]	=	1
  [weight#(x1)]	=	x1 + 0
  [sum(x1, x2)]	=	x2 + 0
  [0]	=	1
  [nil]	=	1
  [weight(x1)]	=	0
  [cons(x1, x2)]	=	x2 + 1
  [sum#(x1, x2)]	=	x1 + 0

  together with the usable rules

  sum(cons(s(n),x),cons(m,y))	→	sum(cons(n,x),cons(s(m),y))	(1)
  sum(nil,y)	→	y	(3)
  sum(cons(0,x),y)	→	sum(x,y)	(2)

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

  sum#(cons(0,x),y)

  →

  sum#(x,y)

  (8)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    sum#(cons(s(n),x),cons(m,y))

    →

    sum#(cons(n,x),cons(s(m),y))

    (6)

    1.1.2.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [s(x1)]	=	x1 + 1
    [weight#(x1)]	=	x1 + 0
    [sum(x1, x2)]	=	x1 + x2 + 0
    [0]	=	44022
    [nil]	=	21680
    [weight(x1)]	=	0
    [cons(x1, x2)]	=	x1 + x2 + 282
    [sum#(x1, x2)]	=	x1 + 0

    together with the usable rules

    sum(cons(s(n),x),cons(m,y))	→	sum(cons(n,x),cons(s(m),y))	(1)
    sum(nil,y)	→	y	(3)
    sum(cons(0,x),y)	→	sum(x,y)	(2)

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

    sum#(cons(s(n),x),cons(m,y))

    →

    sum#(cons(n,x),cons(s(m),y))

    (6)

    could be deleted.

    1.1.2.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.