Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup05)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

1.1 Dependency Pair Transformation

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

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

  →

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

  (10)

  1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 1 with strict dimension 1 over the naturals

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

  together with the usable rules

  b(b(b(b(a(a(b(b(x1))))))))	→	b(b(a(a(b(b(a(a(x1))))))))	(7)
  a(a(a(a(b(b(a(a(x1))))))))	→	a(a(b(b(a(a(b(b(x1))))))))	(8)

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

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

  →

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

  (10)

  could be deleted.

  1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

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

  →

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

  (24)

  1.1.1.1.2 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 1 with strict dimension 1 over the naturals

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

  together with the usable rules

  b(b(b(b(a(a(b(b(x1))))))))	→	b(b(a(a(b(b(a(a(x1))))))))	(7)
  a(a(a(a(b(b(a(a(x1))))))))	→	a(a(b(b(a(a(b(b(x1))))))))	(8)

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

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

  →

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

  (24)

  could be deleted.

  1.1.1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.