Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/multum6)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

1.1 Dependency Pair Transformation

1.1.1 Reduction Pair Processor

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  b#(c(c(x1)))

  →

  b#(c(a(x1)))

  (9)

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

  the pair

  b#(c(c(x1)))

  →

  b#(c(a(x1)))

  (9)

  could be deleted.

  1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  c#(a(b(x1)))	→	a#(c(b(a(x1))))	(15)
  a#(b(c(a(x1))))	→	c#(c(b(x1)))	(12)
  c#(a(b(x1)))	→	c#(b(a(x1)))	(16)
  a#(b(c(a(x1))))	→	c#(b(x1))	(13)

  1.1.1.1.2 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pairs

  c#(a(b(x1)))	→	a#(c(b(a(x1))))	(15)
  c#(a(b(x1)))	→	c#(b(a(x1)))	(16)

  could be deleted.

  1.1.1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.