Certification Problem

Input (TPDB TRS_Innermost/AG01_innermost/#4.17)

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 2 components.

• The 1^st component contains the pair

  f#(g(x),s(0),y)

  →

  f#(g(s(0)),y,g(x))

  (4)

  1.1.1 Usable Rules Processor

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

  g(s(x))	→	s(g(x))	(2)
  g(0)	→	0	(3)

  1.1.1.1 Innermost Lhss Removal Processor

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

  g(s(x0))

  g(0)

  1.1.1.1.1 Rewriting Processor

  We rewrite the right hand side of the pair resulting in

  →

  1.1.1.1.1.1 Usable Rules Processor

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

  g(0)

  →

  0

  (3)

  1.1.1.1.1.1.1 Rewriting Processor

  We rewrite the right hand side of the pair resulting in

  →

  1.1.1.1.1.1.1.1 Usable Rules Processor

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

  There are no rules.

  1.1.1.1.1.1.1.1.1 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.

• The 2^nd component contains the pair

  g#(s(x))

  →

  g#(x)

  (6)

  1.1.2 Usable Rules Processor

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

  There are no rules.

  1.1.2.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.2.1.1 Size-Change Termination

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

  g#(s(x))	→	g#(x)	(6)
  1	>	1

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