Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-18)

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 1 component.

• The 1^st component contains the pair

  c#(y,x,f(z))	→	b#(f(b(z,x)),z)	(8)
  b#(b(y,z),c(a,a,a))	→	c#(z,y,z)	(4)
  c#(y,x,f(z))	→	b#(z,x)	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [b#(x1, x2)]	=	0 · x1 + -8 · x2 + -∞
  [a]	=	4
  [f(x1)]	=	0 · x1 + -∞
  [c#(x1, x2, x3)]	=	-∞ · x1 + 1 · x2 + 1 · x3 + -∞
  [b(x1, x2)]	=	1 · x1 + 1 · x2 + -∞
  [c(x1, x2, x3)]	=	0 · x1 + 2 · x2 + 2 · x3 + 0

  together with the usable rules

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

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

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

  →

  b#(z,x)

  (6)

  could be deleted.

  1.1.1.1 Reduction Pair Processor with Usable Rules

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

  [b#(x1, x2)]	=	1 · x1 + 0 · x2 + 2
  [a]	=	0
  [f(x1)]	=	-2 · x1 + 0
  [c#(x1, x2, x3)]	=	0 · x1 + 1 · x2 + 2 · x3 + 3
  [b(x1, x2)]	=	1 · x1 + 2 · x2 + 2
  [c(x1, x2, x3)]	=	5 · x1 + 1 · x2 + 4 · x3 + 0

  together with the usable rules

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

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

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

  →

  c#(z,y,z)

  (4)

  could be deleted.

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.