Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z29)

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

  a#(a(x,y),z)	→	a#(x,a(y,z))	(13)
  a#(lambda(x),y)	→	a#(y,t)	(11)
  a#(a(x,y),z)	→	a#(y,z)	(10)
  a#(lambda(x),y)	→	a#(x,1)	(9)
  a#(lambda(x),y)	→	a#(x,a(y,t))	(7)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1, x2)]	=	x1 + x2 + 0
  [1]	=	0
  [lambda#(x1)]	=	0
  [t]	=	0
  [lambda(x1)]	=	x1 + 1
  [a#(x1, x2)]	=	x1 + x2 + 0

  together with the usable rules

  lambda(x)	→	x	(4)
  a(lambda(x),y)	→	lambda(a(x,1))	(1)
  a(a(x,y),z)	→	a(x,a(y,z))	(3)
  a(x,y)	→	x	(5)
  a(x,y)	→	y	(6)
  a(lambda(x),y)	→	lambda(a(x,a(y,t)))	(2)

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

  a#(lambda(x),y)	→	a#(y,t)	(11)
  a#(lambda(x),y)	→	a#(x,1)	(9)
  a#(lambda(x),y)	→	a#(x,a(y,t))	(7)

  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

    a#(a(x,y),z)	→	a#(y,z)	(10)
    a#(a(x,y),z)	→	a#(x,a(y,z))	(13)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

    Using the Max-polynomial interpretation

    [a(x1, x2)]	=	x1 + x2 + 1
    [1]	=	0
    [lambda#(x1)]	=	0
    [t]	=	41141
    [lambda(x1)]	=	20163
    [a#(x1, x2)]	=	x1 + 0

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

    a#(a(x,y),z)	→	a#(y,z)	(10)
    a#(a(x,y),z)	→	a#(x,a(y,z))	(13)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.