Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/pair2hard)

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 2 components.

• The 1^st component contains the pair

  p#(a(x0),p(b(a(x1)),x2))	→	p#(a(b(a(x1))),x2)	(4)
  p#(a(x0),p(b(a(x1)),x2))	→	p#(x1,p(a(b(a(x1))),x2))	(5)

  1.1.1 Subterm Criterion Processor

  We use the projection to multisets

  π(p#)	=	{ 2, 2 }
  π(p)	=	{ 2, 2 }

  to remove the pairs:

  p#(a(x0),p(b(a(x1)),x2))

  →

  p#(a(b(a(x1))),x2)

  (4)

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

  [b(x1)]	=	0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 · x1 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  [p#(x1, x2)]	=	1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 · x1 + 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 · x2 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  [a(x1)]	=	0 1 0 0 0 0 1 0 1 0 1 0 0 0 0 1 · x1 + 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
  [p(x1, x2)]	=	1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 · x1 + 1 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 · x2 + 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0

  together with the usable rules

  p(a(x0),p(b(a(x1)),x2))	→	p(x1,p(a(b(a(x1))),x2))	(1)
  a(b(a(x0)))	→	b(a(b(x0)))	(2)

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

  p#(a(x0),p(b(a(x1)),x2))

  →

  p#(x1,p(a(b(a(x1))),x2))

  (5)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  a#(b(a(x0)))

  →

  a#(b(x0))

  (6)

  1.1.2 Subterm Criterion Processor

  We use the projection to multisets

  π(a#)	=	{ 1 }
  π(b)	=	{ 1 }

  to remove the pairs:

  a#(b(a(x0)))

  →

  a#(b(x0))

  (6)

  1.1.2.1 P is empty

  There are no pairs anymore.