Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/9)

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

  a#(x,y)	→	c#(y)	(6)
  c#(b(y,c(x)))	→	c#(c(b(a(0,0),y)))	(7)
  c#(b(y,c(x)))	→	c#(b(a(0,0),y))	(8)
  c#(b(y,c(x)))	→	a#(0,0)	(10)

  1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  a#(0,0)

  →

  c#(0)

  (11)

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    c#(b(y,c(x)))	→	c#(b(a(0,0),y))	(8)
    c#(b(y,c(x)))	→	c#(c(b(a(0,0),y)))	(7)

    1.1.1.1.1 Reduction Pair Processor

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

    [c#(x1)]	=	-∞ + 0 · x1
    [b(x1, x2)]	=	1 + 0 · x1 + -∞ · x2
    [c(x1)]	=	0 + -∞ · x1
    [a(x1, x2)]	=	1 + 1 · x1 + -∞ · x2
    [0]	=	0

    the pair

    c#(b(y,c(x)))

    →

    c#(c(b(a(0,0),y)))

    (7)

    could be deleted.

    1.1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the rationals with delta = 3

    [c#(x1)]	=	0 + 4 · x1
    [b(x1, x2)]	=	0 + 1 · x1 + 1/4 · x2
    [c(x1)]	=	4 + 0 · x1
    [a(x1, x2)]	=	1/4 + 4 · x1 + 0 · x2
    [0]	=	0

    the pair

    c#(b(y,c(x)))

    →

    c#(b(a(0,0),y))

    (8)

    could be deleted.

    1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.