Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-22)

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 2 components.

• The 1^st component contains the pair

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

  →

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

  (7)

  1.1.1 Reduction Pair Processor

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

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

  the pair

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

  →

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

  (7)

  could be deleted.

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  b#(a,b(c(z,x,y),a))	→	c#(y,z,a)	(6)
  c#(f(c(a,y,a)),x,z)	→	b#(y,b(x,a))	(14)

  1.1.2 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

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

  1.1.2.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

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

  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

  b#(a,b(c(z,x,y),a))	→	c#(y,z,a)	(6)
  c#(f(c(a,y,a)),x,z)	→	b#(y,b(x,a))	(14)

  and no rules could be deleted.

  1.1.2.1.1 P is empty

  There are no pairs anymore.