Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-15)

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

  c#(z,x,a)	→	b#(b(f(z),z),x)	(5)
  b#(y,b(z,a))	→	b#(c(f(a),y,z),z)	(9)
  b#(y,b(z,a))	→	c#(f(a),y,z)	(10)
  c#(z,x,a)	→	b#(f(z),z)	(6)

  1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  c#(f(a),y_0,a)

  →

  b#(b(f(f(a)),f(a)),y_0)

  (12)

  1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  c#(f(a),y_0,a)

  →

  b#(f(f(a)),f(a))

  (13)

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    b#(y,b(z,a))	→	c#(f(a),y,z)	(10)
    c#(f(a),y_0,a)	→	b#(b(f(f(a)),f(a)),y_0)	(12)
    b#(y,b(z,a))	→	b#(c(f(a),y,z),z)	(9)

    1.1.1.1.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

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

    the pairs

    c#(f(a),y_0,a)	→	b#(b(f(f(a)),f(a)),y_0)	(12)
    b#(y,b(z,a))	→	b#(c(f(a),y,z),z)	(9)

    could be deleted.

    1.1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

    Using the linear polynomial interpretation over the naturals

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

    having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

    1.1.1.1.1.1.1.1 Size-Change Termination

    Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

    b#(y,b(z,a))	→	c#(f(a),y,z)	(10)
    1	≥	2
    2	>	3

    As there is no critical graph in the transitive closure, there are no infinite chains.