Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-289)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Reduction Pair Processor

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  b#(a(b(b(x1))))	→	b#(a(b(a(x1))))	(15)
  b#(b(a(b(x1))))	→	b#(b(b(b(x1))))	(9)

  1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

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

  the pair

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

  →

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

  (9)

  could be deleted.

  1.1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

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

  the pair

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

  →

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

  (15)

  could be deleted.

  1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

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

  →

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

  (5)

  1.1.1.2 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  together with the usable rules

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

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

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

  →

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

  (5)

  could be deleted.

  1.1.1.2.1 P is empty

  There are no pairs anymore.