Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-28)

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#(y,z)	→	c#(y,z,z)	(5)
  c#(f(z),f(c(a,x,a)),y)	→	c#(f(b(x,z)),c(z,y,a),a)	(6)
  c#(f(z),f(c(a,x,a)),y)	→	b#(x,z)	(7)
  c#(f(z),f(c(a,x,a)),y)	→	c#(z,y,a)	(8)

  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, x2)]	=	2 -∞ -∞ + 1 1 1 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 0 1 0 -∞ -∞ -∞ -∞ -∞ -∞ · x2
  [c#(x1, x2, x3)]	=	0 -∞ -∞ + 0 -∞ 0 -∞ -∞ -∞ -∞ -∞ -∞ · x1 + -∞ 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x2 + -∞ 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x3
  [f(x1)]	=	2 0 0 + 0 1 0 1 -∞ 0 -∞ -∞ -∞ · x1
  [c(x1, x2, x3)]	=	0 -∞ -∞ + -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 0 0 0 -∞ 0 -∞ 0 0 0 · x2 + 0 0 0 0 0 0 0 0 0 · x3
  [a]	=	0 0 0
  [b(x1, x2)]	=	2 1 0 + 0 0 1 0 0 0 0 0 0 · x1 + -∞ 0 0 -∞ -∞ -∞ -∞ 1 0 · x2

  the pair

  b#(y,z)

  →

  c#(y,z,z)

  (5)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    c#(f(z),f(c(a,x,a)),y)	→	c#(z,y,a)	(8)
    c#(f(z),f(c(a,x,a)),y)	→	c#(f(b(x,z)),c(z,y,a),a)	(6)

    1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

    c#(f(x0),f(c(a,x1,a)),a)

    →

    c#(x0,a,a)

    (9)

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      c#(f(z),f(c(a,x,a)),y)

      →

      c#(f(b(x,z)),c(z,y,a),a)

      (6)

      1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      c#(f(y_2),f(c(a,x1,a)),a)

      →

      c#(f(b(x1,y_2)),c(y_2,a,a),a)

      (10)

      1.1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      There are no rules.

      1.1.1.1.1.1.1.1.1 P is empty

      There are no pairs anymore.