Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/aabaabbaab-aabbaabaabbaa.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#(a(b(a(a(b(b(a(a(b(x1))))))))))	→	a#(a(b(a(a(b(b(a(a(x1)))))))))	(7)
  a#(a(b(a(a(b(b(a(a(b(x1))))))))))	→	a#(a(x1))	(3)
  a#(a(b(a(a(b(b(a(a(b(x1))))))))))	→	a#(x1)	(2)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers

  [a(x1)]	=	-∞ 0 0 -∞ · x1 + 2 -∞ 2 -∞
  [b(x1)]	=	0 -∞ 1 0 · x1 + 0 -∞ 0 -∞
  [a#(x1)]	=	0 -∞ -∞ -∞ · x1 + 0 -∞ -∞ -∞

  together with the usable rule

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

  →

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

  (1)

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

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

  →

  a#(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

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

  together with the usable rule

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

  →

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

  (1)

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

  a#(a(b(a(a(b(b(a(a(b(x1))))))))))	→	a#(a(b(a(a(b(b(a(a(x1)))))))))	(7)
  a#(a(b(a(a(b(b(a(a(b(x1))))))))))	→	a#(a(x1))	(3)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.