Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/10)

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(c(y)))	→	c#(a(y,0))	(5)
  c#(a(a(0,x),y))	→	c#(c(c(0)))	(6)
  c#(c(c(y)))	→	c#(c(a(y,0)))	(4)
  c#(a(a(0,x),y))	→	c#(c(0))	(7)

  1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

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

  the pair

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

  →

  c#(c(0))

  (7)

  could be deleted.

  1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 1 to get the following set of pairs

  c#(c(c(a(0,x0))))	→	c#(a(c(c(c(0))),0))	(9)
  c#(c(c(y)))	→	c#(a(y,0))	(5)

  1.1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [c#(x1)]	=	-2 + x1
  [c(x1)]	=	1 + x1
  [a(x1, x2)]	=	1
  [0]	=	0

  the pair

  c#(c(c(a(0,x0))))

  →

  c#(a(c(c(c(0))),0))

  (9)

  could be deleted.

  1.1.1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

  [c#(x1)]	=	0 0 + 0 2 0 0 · x1
  [c(x1)]	=	1 0 + 1 0 2 2 · x1
  [a(x1, x2)]	=	1 0 + 0 0 2 0 · x1 + 0 0 0 0 · x2
  [0]	=	0 0

  the pair

  c#(c(c(y)))

  →

  c#(a(y,0))

  (5)

  could be deleted.

  1.1.1.1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 1 to get the following set of pairs

  c#(a(a(0,y0),y1))

  →

  c#(c(0))

  (10)

  1.1.1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  together with the usable rule

  c(y)

  →

  y

  (3)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.1.1.1.1.1.1.1 Switch to Innermost Termination

  The TRS does not have overlaps with the pairs and is locally confluent:

  20

  Hence, it suffices to show innermost termination in the following.

  1.1.1.1.1.1.1.1.1.1 Rewriting Processor

  We rewrite the right hand side of the pair resulting in

  →

  1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

  We restrict the rewrite rules to the following usable rules of the DP problem.

  There are no rules.

  1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

  We restrict the innermost strategy to the following left hand sides.

  There are no lhss.

  1.1.1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  There are no rules.

  1.1.1.1.1.1.1.1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.