Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/abaabaaaa-aaaaabaababaab.srs)

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

  a#(b(a(a(b(a(a(a(a(x1)))))))))	→	a#(b(a(a(b(a(b(a(a(b(x1))))))))))	(7)
  a#(b(a(a(b(a(a(a(a(x1)))))))))	→	a#(b(a(a(b(x1)))))	(4)
  a#(b(a(a(b(a(a(a(a(x1)))))))))	→	a#(b(x1))	(2)

  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

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

  together with the usable rule

  a(b(a(a(b(a(a(a(a(x1)))))))))

  →

  a(a(a(a(a(b(a(a(b(a(b(a(a(b(x1))))))))))))))

  (1)

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

  a#(b(a(a(b(a(a(a(a(x1)))))))))

  →

  a#(b(x1))

  (2)

  could be deleted.

  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 1 0 0 0 0 0 0 · x1 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  [a(x1)]	=	0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 · x1 + 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
  [a#(x1)]	=	1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 · x1 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

  together with the usable rule

  a(b(a(a(b(a(a(a(a(x1)))))))))

  →

  a(a(a(a(a(b(a(a(b(a(b(a(a(b(x1))))))))))))))

  (1)

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

  a#(b(a(a(b(a(a(a(a(x1)))))))))	→	a#(b(a(a(b(a(b(a(a(b(x1))))))))))	(7)
  a#(b(a(a(b(a(a(a(a(x1)))))))))	→	a#(b(a(a(b(x1)))))	(4)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.