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 AProVE @ 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)	(7)
  f#(f(f(c(z,x,a))))	→	b#(f(x),z)	(8)
  b#(y,b(a,z))	→	b#(f(c(y,y,a)),b(f(z),a))	(4)
  f#(f(f(c(z,x,a))))	→	f#(x)	(9)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

  prec(c)	=	1		weight(c)	=	1
  prec(a)	=	0		weight(a)	=	1

  in combination with the following argument filter

  π(b#)	=	2
  π(b)	=	2
  π(f#)	=	1
  π(f)	=	1
  π(c)	=	[1,2]
  π(a)	=	[]

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

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

  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))

    (4)

    1.1.1.1.1 Reduction Pair Processor

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

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

    the pair

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

    →

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

    (4)

    could be deleted.

    1.1.1.1.1.1 P is empty

    There are no pairs anymore.