Certification Problem

Input (TPDB SRS_Standard/Trafo_06/un09)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ 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

  b#(a(b(b(x1))))	→	b#(b(a(b(a(b(x1))))))	(5)
  b#(b(b(a(b(x1)))))	→	b#(a(b(b(a(b(x1))))))	(4)
  b#(a(b(b(x1))))	→	b#(a(b(a(b(x1)))))	(6)
  b#(a(b(b(x1))))	→	b#(a(b(x1)))	(7)
  b#(a(b(a(a(b(b(x1)))))))	→	b#(b(b(a(b(x1)))))	(10)
  b#(a(b(a(a(b(b(x1)))))))	→	b#(b(a(b(x1))))	(11)
  b#(a(b(a(a(b(b(x1)))))))	→	b#(a(b(x1)))	(12)

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

  the pairs

  b#(a(b(a(a(b(b(x1)))))))	→	b#(b(b(a(b(x1)))))	(10)
  b#(a(b(a(a(b(b(x1)))))))	→	b#(b(a(b(x1))))	(11)
  b#(a(b(a(a(b(b(x1)))))))	→	b#(a(b(x1)))	(12)

  could be deleted.

  1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pairs

  b#(b(b(a(b(x1)))))	→	b#(a(b(b(a(b(x1))))))	(4)
  b#(a(b(b(x1))))	→	b#(a(b(a(b(x1)))))	(6)
  b#(a(b(b(x1))))	→	b#(a(b(x1)))	(7)

  could be deleted.

  1.1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pair

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

  →

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

  (5)

  could be deleted.

  1.1.1.1.1.1 P is empty

  There are no pairs anymore.