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

  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 arctic semiring over the integers

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

  together with the usable rules

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

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

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

  →

  c#(z,y,x)

  (4)

  could be deleted.

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

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

  together with the usable rules

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

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

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

  →

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

  (6)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.