Certification Problem

Input (TPDB TRS_Standard/SK90/2.61)

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

  f#(j(x,y),y)

  →

  f#(x,k(y))

  (14)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [h(x1)]	=	1
  [s(x1)]	=	1
  [k(x1)]	=	28959
  [h2#(x1, x2, x3)]	=	0
  [f(x1, x2)]	=	0
  [0]	=	28958
  [k#(x1)]	=	0
  [f#(x1, x2)]	=	x1 + x2 + 0
  [g#(x1)]	=	0
  [j(x1, x2)]	=	x1 + 28960
  [h1(x1, x2)]	=	x1 + 1
  [i(x1)]	=	0
  [h2(x1, x2, x3)]	=	0
  [g(x1)]	=	0
  [i#(x1)]	=	0

  together with the usable rules

  k(h1(x,y))	→	h1(s(x),y)	(8)
  k(h(x))	→	h1(0,x)	(7)

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

  f#(j(x,y),y)

  →

  f#(x,k(y))

  (14)

  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

  h2#(x,j(y,h1(z,u)),h1(z,u))

  →

  h2#(s(x),y,h1(s(z),u))

  (10)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [h(x1)]	=	1
  [s(x1)]	=	1
  [k(x1)]	=	28959
  [h2#(x1, x2, x3)]	=	x1 + x2 + x3 + 0
  [f(x1, x2)]	=	0
  [0]	=	28958
  [k#(x1)]	=	0
  [f#(x1, x2)]	=	x2 + 0
  [g#(x1)]	=	0
  [j(x1, x2)]	=	x1 + x2 + 2
  [h1(x1, x2)]	=	x1 + 1
  [i(x1)]	=	0
  [h2(x1, x2, x3)]	=	0
  [g(x1)]	=	0
  [i#(x1)]	=	0

  together with the usable rules

  k(h1(x,y))	→	h1(s(x),y)	(8)
  k(h(x))	→	h1(0,x)	(7)

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

  h2#(x,j(y,h1(z,u)),h1(z,u))

  →

  h2#(s(x),y,h1(s(z),u))

  (10)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.