Certification Problem

Input (TPDB TRS_Standard/Waldmann_06/jwtpa2)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  f#(x,f(a,y))	→	f#(f(f(a,x),h(a)),y)	(4)
  f#(x,f(a,y))	→	f#(a,f(f(f(a,x),h(a)),y))	(5)
  f#(x,f(a,y))	→	f#(a,x)	(2)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [f(x1, x2)]	=	0 · x1 + 4 · x2 + 4
  [f#(x1, x2)]	=	4 · x1 + 4 · x2 + 0
  [a]	=	0
  [h(x1)]	=	0 · x1 + 0

  together with the usable rule

  f(x,f(a,y))

  →

  f(a,f(f(f(a,x),h(a)),y))

  (1)

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

  f#(x,f(a,y))

  →

  f#(a,x)

  (2)

  could be deleted.

  1.1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(f#)	=	{ 2, 2 }
  π(f)	=	{ 2, 2 }

  to remove the pairs:

  f#(x,f(a,y))

  →

  f#(f(f(a,x),h(a)),y)

  (4)

  1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

  [f(x1, x2)]	=	0 0 0 1 0 0 0 1 0 · x1 + 0 0 0 0 0 1 0 1 0 · x2 + 0 0 0 0 0 0 1 0 0
  [f#(x1, x2)]	=	0 0 1 0 0 0 0 0 0 · x1 + 0 1 0 0 0 0 0 0 0 · x2 + 0 0 0 0 0 0 0 0 0
  [a]	=	1 0 0 1 0 0 0 0 0
  [h(x1)]	=	0 0 0 0 1 0 0 0 0 · x1 + 0 0 0 0 0 0 0 0 0

  together with the usable rule

  f(x,f(a,y))

  →

  f(a,f(f(f(a,x),h(a)),y))

  (1)

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

  f#(x,f(a,y))

  →

  f#(a,f(f(f(a,x),h(a)),y))

  (5)

  could be deleted.

  1.1.1.1.1.1 P is empty

  There are no pairs anymore.