Certification Problem

Input (TPDB TRS_Standard/SK90/4.55)

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,x)

  →

  f#(a,b)

  (3)

  1.1.1.1 Usable Rules Processor

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

  b

  →

  c

  (2)

  1.1.1.1.1 Innermost Lhss Removal Processor

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

  b

  1.1.1.1.1.1 Rewriting Processor

  We rewrite the right hand side of the pair resulting in

  →

  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 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.1.1.1.1.1.1.1 Instantiation Processor

  We instantiate the pair to the following set of pairs

  There are no rules.

  1.1.1.1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.