Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove01)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.

• The 1^st component contains the pair

  j#(s(x1))	→	f#(p(s(p(p(s(x1))))))	(23)
  f#(s(x1))	→	g#(p(s(s(x1))))	(13)
  g#(s(x1))	→	j#(s(p(s(p(s(x1))))))	(17)

  1.1.1 Reduction Pair Processor with Usable Rules

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

  [p(x1)]	=	-3 · x1 + 0
  [g#(x1)]	=	-2 · x1 + 0
  [0(x1)]	=	-∞ · x1 + 0
  [j#(x1)]	=	-3 · x1 + 0
  [f#(x1)]	=	0 · x1 + 0
  [s(x1)]	=	3 · x1 + 2

  together with the usable rules

  p(s(x1))	→	x1	(2)
  p(0(x1))	→	0(s(s(p(x1))))	(1)
  p(p(s(x1)))	→	p(x1)	(3)

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

  f#(s(x1))

  →

  g#(p(s(s(x1))))

  (13)

  could be deleted.

  1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 0 components.

• The 2^nd component contains the pair

  half#(s(s(x1)))	→	half#(p(p(s(s(x1)))))	(31)
  half#(0(x1))	→	half#(p(s(p(s(x1)))))	(28)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the

  prec(half#)	=	0		stat(half#)	=	lex
  prec(s)	=	0		stat(s)	=	lex
  prec(p)	=	0		stat(p)	=	lex
  prec(0)	=	0		stat(0)	=	lex

  π(half#)	=	1
  π(s)	=	1
  π(p)	=	1
  π(0)	=	[1]

  together with the usable rules

  p(s(x1))	→	x1	(2)
  p(0(x1))	→	0(s(s(p(x1))))	(1)
  p(p(s(x1)))	→	p(x1)	(3)

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

  half#(0(x1))

  →

  half#(p(s(p(s(x1)))))

  (28)

  could be deleted.

  1.1.2.1 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)]	=	1 · x1 + 0
  [s(x1)]	=	1 · x1 + 0

  together with the usable rules

  p(s(x1))	→	x1	(2)
  p(0(x1))	→	0(s(s(p(x1))))	(1)
  p(p(s(x1)))	→	p(x1)	(3)

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

  half#(s(s(x1)))

  →

  half#(p(p(s(s(x1)))))

  (31)

  could be deleted.

  1.1.2.1.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  p#(p(s(x1)))	→	p#(x1)	(11)
  p#(0(x1))	→	p#(x1)	(10)

  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)	(11)
  1	>	1
  p#(0(x1))	→	p#(x1)	(10)
  1	>	1

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

• The 4^th component contains the pair

  rd#(0(x1))

  →

  rd#(x1)

  (32)

  1.1.4 Size-Change Termination

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

  rd#(0(x1))	→	rd#(x1)	(32)
  1	>	1

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