Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove07)

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

  foo#(s(x1))	→	bar#(p(p(s(s(p(s(x1)))))))	(22)
  bar#(s(x1))	→	foo#(s(p(p(s(s(x1))))))	(30)
  foo#(s(x1))	→	foo#(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))	(18)

  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 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(x1)))))	(7)
  bar(0(x1))	→	0(p(s(s(s(x1)))))	(3)
  bar(s(x1))	→	p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))	(4)
  foo(s(x1))	→	p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))	(2)
  foo(0(x1))	→	0(s(p(p(p(s(s(s(p(s(x1))))))))))	(1)

  1.1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [bar#(x1)]	=	2 + x1
  [foo#(x1)]	=	2 + x1
  [p(x1)]	=	-2 + x1
  [s(x1)]	=	2 + x1
  [0(x1)]	=	-2
  [bar(x1)]	=	1 + x1
  [foo(x1)]	=	2

  the pairs

  foo#(s(x1))	→	bar#(p(p(s(s(p(s(x1)))))))	(22)
  foo#(s(x1))	→	foo#(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))	(18)

  could be deleted.

  1.1.1.1.1.1 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(x1)))))	(7)

  1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

  p(s(x0))

  p(0(x0))

  1.1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [bar#(x1)]	=	1 + 1 · x1
  [s(x1)]	=	0
  [foo#(x1)]	=	0
  [p(x1)]	=	1 · x1
  [0(x1)]	=	1 + 1 · x1

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

  bar#(s(x1))

  →

  foo#(s(p(p(s(s(x1))))))

  (30)

  could be deleted.

  1.1.1.1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  p#(p(s(x1)))

  →

  p#(x1)

  (33)

  1.1.2 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.

  1.1.2.1 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)	(33)
  1	>	1

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