Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-25)

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

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

  1.1.1 Reduction Pair Processor

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

  [c#(x1, x2, x3)]	=	-∞ -∞ -∞ + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x1 + -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x2 + 1 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x3
  [c(x1, x2, x3)]	=	0 0 -∞ + 0 -∞ -∞ -∞ -∞ -∞ -∞ -∞ -∞ · x1 + 1 -∞ -∞ -∞ -∞ -∞ 0 1 0 · x2 + -∞ 0 3 -∞ -∞ -∞ 0 -∞ -∞ · x3
  [a]	=	0 -∞ 0
  [f(x1)]	=	0 -∞ -∞ + -∞ 0 -∞ -∞ -∞ 0 -∞ -∞ 0 · x1
  [b(x1, x2)]	=	-∞ -∞ -∞ + 0 -∞ -∞ 0 0 0 0 -∞ -∞ · x1 + 0 -∞ -∞ 0 -∞ 0 0 -∞ -∞ · x2

  the pair

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

  →

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

  (6)

  could be deleted.

  1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [c(x1, x2, x3)]	=	1 · x1 + 1 · x2 + 1 · x3
  [a]	=	0
  [c#(x1, x2, x3)]	=	1 · x1 + 1 · x2 + 1 · x3

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.1.1.1 Size-Change Termination

  Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

  c#(c(z,x,a),a,y)	→	c#(z,y,x)	(8)
  1	>	1
  3	≥	2
  1	>	3

  As there is no critical graph in the transitive closure, there are no infinite chains.