Certification Problem

Input (TPDB TRS_Innermost/Mixed_innermost/innermost4)

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

  f#(a(x),y,s(z),u)	→	g#(x,y,s(z),u)	(7)
  g#(x,y,z,u)	→	h#(x,y,z,u)	(8)
  h#(b,y,z,u)	→	f#(y,y,z,u)	(9)
  f#(a(x),y,s(z),u)	→	f#(a(b),y,z,g(x,y,s(z),u))	(5)

  1.1.1 Rewriting Processor

  We rewrite the right hand side of the pair resulting in

  →

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    g#(x,y,z,u)	→	h#(x,y,z,u)	(8)
    h#(b,y,z,u)	→	f#(y,y,z,u)	(9)
    f#(a(x),y,s(z),u)	→	g#(x,y,s(z),u)	(7)

    1.1.1.1.1 Usable Rules Processor

    We restrict the rewrite rules to the following usable rules of the DP problem.

    There are no rules.

    1.1.1.1.1.1 Innermost Lhss Removal Processor

    We restrict the innermost strategy to the following left hand sides.

    a(b)

    1.1.1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

    f#(a(x0),a(x0),s(x2),z2)

    →

    g#(x0,a(x0),s(x2),z2)

    (11)

    1.1.1.1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

    g#(z0,a(z0),s(z1),z2)

    →

    h#(z0,a(z0),s(z1),z2)

    (12)

    1.1.1.1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 0 components.