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 ttt2 @ 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))

  (9)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals

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

  together with the usable rules

  b(a,b(c(z,x,y),a))	→	b(b(z,c(y,z,a)),x)	(1)
  f(c(a,b(b(z,a),y),x))	→	f(c(x,b(z,x),y))	(2)
  c(f(c(a,y,a)),x,z)	→	f(b(b(z,z),f(b(y,b(x,a)))))	(3)

  (w.r.t. the implicit argument filter of the reduction pair), the pair

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

  →

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

  (9)

  could be deleted.

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  c#(f(c(a,y,a)),x,z)	→	b#(y,b(x,a))	(11)
  b#(a,b(c(z,x,y),a))	→	c#(y,z,a)	(4)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the arctic semiring over the integers

  [b#(x1, x2)]	=	0 · x1 + 0 · x2 + 0
  [c(x1, x2, x3)]	=	3 · x1 + 0 · x2 + 0 · x3 + 1
  [f(x1)]	=	0 · x1 + 0
  [c#(x1, x2, x3)]	=	0 · x1 + 4 · x2 + 0 · x3 + 0
  [a]	=	0
  [b(x1, x2)]	=	4 · x1 + 0 · x2 + 3

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

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

  →

  c#(y,z,a)

  (4)

  could be deleted.

  1.1.2.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.