Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/8)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by NaTT @ 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#(c(a(a(c(b(0,y)),0),0)))	(9)
  c#(c(c(y)))	→	c#(a(a(c(b(0,y)),0),0))	(4)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a(x1, x2)]	=	x1 + 0
  [b(x1, x2)]	=	x1 + x2 + 0
  [c(x1)]	=	x1 + 20163
  [0]	=	0
  [c#(x1)]	=	x1 + 0
  [a#(x1, x2)]	=	0
  [b#(x1, x2)]	=	0

  together with the usable rules

  b(b(0,y),x)	→	y	(1)
  a(y,0)	→	b(y,0)	(3)
  c(c(c(y)))	→	c(c(a(a(c(b(0,y)),0),0)))	(2)

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

  c#(c(c(y)))

  →

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

  (4)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    c#(c(c(y)))

    →

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

    (9)

    1.1.1.1.1 Reduction Pair Processor with Usable Rules

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

    [a(x1, x2)]	=	1 0 1 0 · x1 + 0 0 1 0 · x2 + 17889 0
    [b(x1, x2)]	=	1 0 1 0 · x1 + 1 1 0 0 · x2 + 0 1
    [c(x1)]	=	1 1 1 1 · x1 + 1 53676
    [0]	=	1 17888
    [c#(x1)]	=	1 1 1 1 · x1 + 0 0
    [a#(x1, x2)]	=	0 0
    [b#(x1, x2)]	=	0 0

    together with the usable rules

    b(b(0,y),x)	→	y	(1)
    a(y,0)	→	b(y,0)	(3)
    c(c(c(y)))	→	c(c(a(a(c(b(0,y)),0),0)))	(2)

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

    c#(c(c(y)))

    →

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

    (9)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.