Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/aprove2)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

1.1 Rule Removal

1.1.1 Rule Removal

1.1.1.1 Rule Removal

1.1.1.1.1 Dependency Pair Transformation

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  s#(s(f(f(x1))))	→	s#(s(s(x1)))	(23)
  s#(s(f(f(x1))))	→	s#(s(x1))	(22)
  s#(s(f(f(x1))))	→	s#(x1)	(21)

  1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

  [g(x1)]	=	0 1 0 0 1 1 0 0 0 · x1 + 1 0 0 1 0 0 0 0 0
  [f(x1)]	=	1 0 1 0 0 0 0 1 0 · x1 + 1 0 0 0 0 0 0 0 0
  [h(x1)]	=	1 0 0 1 1 0 0 0 1 · x1 + 0 0 0 0 0 0 0 0 0
  [s#(x1)]	=	0 1 0 0 0 0 0 0 0 · x1 + 0 0 0 0 0 0 0 0 0
  [s(x1)]	=	0 1 0 1 0 0 0 0 0 · x1 + 0 0 0 0 0 0 0 0 0

  together with the usable rules

  h(g(x1))	→	s(f(g(x1)))	(9)
  s(s(s(f(x1))))	→	h(s(f(h(x1))))	(10)
  h(f(x1))	→	h(s(f(h(x1))))	(11)
  h(x1)	→	x1	(4)
  s(s(f(f(x1))))	→	f(f(s(s(s(x1)))))	(12)

  (w.r.t. the implicit argument filter of the reduction pair), the pairs

  s#(s(f(f(x1))))	→	s#(s(s(x1)))	(23)
  s#(s(f(f(x1))))	→	s#(s(x1))	(22)
  s#(s(f(f(x1))))	→	s#(x1)	(21)

  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

  h#(f(x1))

  →

  h#(x1)

  (18)

  1.1.1.1.1.1.2 Size-Change Termination

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

  h#(f(x1))	→	h#(x1)	(18)
  1	>	1

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