Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-10)

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

  b#(y,b(a,z))	→	f#(z)	(9)
  b#(y,b(a,z))	→	b#(f(c(y,y,a)),b(f(z),a))	(8)
  f#(f(f(c(z,x,a))))	→	f#(x)	(5)
  f#(f(f(c(z,x,a))))	→	b#(f(x),z)	(6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Max-polynomial interpretation

  [a]	=	0
  [b(x1, x2)]	=	x2 + 28105
  [c(x1, x2, x3)]	=	x1 + x2 + 1
  [f(x1)]	=	x1 + 1
  [f#(x1)]	=	x1 + 0
  [b#(x1, x2)]	=	x2 + 2

  together with the usable rule

  b(y,b(a,z))

  →

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

  (2)

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

  b#(y,b(a,z))	→	f#(z)	(9)
  f#(f(f(c(z,x,a))))	→	f#(x)	(5)
  f#(f(f(c(z,x,a))))	→	b#(f(x),z)	(6)

  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

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

    →

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

    (8)

    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]	=	1 0
    [b(x1, x2)]	=	1 0 1 0 · x1 + 0 1 1 1 · x2 + 27233 0
    [c(x1, x2, x3)]	=	0 1 1 0 · x1 + 0 1 0 0 · x2 + 36242 0
    [f(x1)]	=	0 1 1 1 · x1 + 0 1
    [f#(x1)]	=	0 0
    [b#(x1, x2)]	=	0 0 1 0 · x1 + 1 0 0 0 · x2 + 0 0

    together with the usable rules

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

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

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

    →

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

    (8)

    could be deleted.

    1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.