Certification Problem

Input (TPDB TRS_Standard/Zantema_05/jw19)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

• The 1^st component contains the pair

  f#(x,f(a,a))	→	f#(a,x)	(4)
  f#(x,f(a,a))	→	f#(f(f(a,a),a),f(a,x))	(2)

  1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  f#(a,f(a,a))	→	f#(a,a)	(5)
  f#(f(f(a,a),a),f(a,a))	→	f#(a,f(f(a,a),a))	(6)

  1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    f#(x,f(a,a))

    →

    f#(f(f(a,a),a),f(a,x))

    (2)

    1.1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

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

    →

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

    (7)

    1.1.1.1.1.1.1 Usable Rules Processor

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

    There are no rules.

    1.1.1.1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

    There are no rules.

    1.1.1.1.1.1.1.1.1 P is empty

    There are no pairs anymore.