Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove03)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ 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

  half#(s(s(x1)))	→	half#(p(p(s(s(p(s(x1)))))))	(24)
  half#(s(x1))	→	half#(p(p(s(s(p(s(x1)))))))	(15)

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the arctic semiring over the integers

  [p(x1)]	=	-1 · x1 + 0
  [0(x1)]	=	-∞ · x1 + 0
  [half#(x1)]	=	0 · x1 + 0
  [s(x1)]	=	1 · x1 + 1

  together with the usable rules

  p(s(x1))	→	x1	(9)
  p(p(s(x1)))	→	p(x1)	(8)
  p(0(x1))	→	0(s(s(s(s(x1)))))	(10)

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

  half#(s(s(x1)))	→	half#(p(p(s(s(p(s(x1)))))))	(24)
  half#(s(x1))	→	half#(p(p(s(s(p(s(x1)))))))	(15)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  sixtimes#(s(x1))

  →

  sixtimes#(p(s(p(p(p(s(s(s(x1)))))))))

  (33)

  1.1.1.2 Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the arctic semiring over the integers

  [sixtimes#(x1)]	=	0 · x1 + 0
  [p(x1)]	=	-1 · x1 + 0
  [0(x1)]	=	-∞ · x1 + 0
  [s(x1)]	=	1 · x1 + 1

  together with the usable rules

  p(s(x1))	→	x1	(9)
  p(p(s(x1)))	→	p(x1)	(8)
  p(0(x1))	→	0(s(s(s(s(x1)))))	(10)

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

  sixtimes#(s(x1))

  →

  sixtimes#(p(s(p(p(p(s(s(s(x1)))))))))

  (33)

  could be deleted.

  1.1.1.2.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  p#(p(s(x1)))

  →

  p#(x1)

  (39)

  1.1.1.3 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)	(39)
  1	>	1

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