Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove04)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.

• The 1^st component contains the pair

  sq#(s(x1))

  →

  sq#(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

  (19)

  1.1.1.1 Switch to Innermost Termination

  The TRS does not have overlaps with the pairs and is locally confluent:

  20

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

  1.1.1.1.1 Usable Rules Processor

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

  p(s(x1))	→	x1	(6)
  p(0(x1))	→	0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))	(7)

  1.1.1.1.1.1 Innermost Lhss Removal Processor

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

  p(s(x0))

  p(0(x0))

  1.1.1.1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [sq#(x1)]	=	2 + 2 · x1
  [p(x1)]	=	-2 + x1
  [s(x1)]	=	2 + x1
  [0(x1)]	=	-2

  the pair

  sq#(s(x1))

  →

  sq#(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

  (19)

  could be deleted.

  1.1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  twice#(s(x1))

  →

  twice#(p(s(p(s(x1)))))

  (43)

  1.1.1.2 Switch to Innermost Termination

  The TRS does not have overlaps with the pairs and is locally confluent:

  20

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

  1.1.1.2.1 Usable Rules Processor

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

  p(s(x1))

  →

  x1

  (6)

  1.1.1.2.1.1 Innermost Lhss Removal Processor

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

  p(s(x0))

  p(0(x0))

  1.1.1.2.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [twice#(x1)]	=	2 + x1
  [p(x1)]	=	-1 + x1
  [s(x1)]	=	1 + x1

  the pair

  twice#(s(x1))

  →

  twice#(p(s(p(s(x1)))))

  (43)

  could be deleted.

  1.1.1.2.1.1.1.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  p#(p(s(x1)))

  →

  p#(x1)

  (46)

  1.1.1.3 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [p(x1)]	=	1 · x1
  [s(x1)]	=	1 · x1
  [p#(x1)]	=	1 · x1

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

  1.1.1.3.1 Size-Change Termination

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

  p#(p(s(x1)))	→	p#(x1)	(46)
  1	>	1

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