Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/aprove1)

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 4 components.

• The 1^st component contains the pair

  i#(x1)	→	f#(p(x1))	(17)
  f#(s(x1))	→	g#(s(x1))	(14)
  g#(x1)	→	i#(s(half(x1)))	(15)

  1.1.1 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [half(x1)]	=	1 · x1
  [0(x1)]	=	1 · x1
  [s(x1)]	=	1 · x1
  [p(x1)]	=	1 · x1
  [f#(x1)]	=	1 · x1
  [i#(x1)]	=	1 · x1
  [g#(x1)]	=	1 · x1

  together with the usable rules

  half(0(x1))	→	0(s(s(half(x1))))	(7)
  half(s(s(x1)))	→	s(half(p(p(s(s(x1))))))	(8)
  p(s(x1))	→	x1	(2)
  p(0(x1))	→	0(s(s(p(x1))))	(1)
  p(p(s(x1)))	→	p(x1)	(3)
  0(x1)	→	x1	(9)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.1.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [half(x1)]	=	1 · x1
  [0(x1)]	=	2 + 1 · x1
  [s(x1)]	=	1 · x1
  [p(x1)]	=	1 · x1
  [i#(x1)]	=	1 · x1
  [f#(x1)]	=	1 · x1
  [g#(x1)]	=	1 · x1

  the rule

  0(x1)

  →

  x1

  (9)

  could be deleted.

  1.1.1.1.1 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.

  1.1.1.1.1.1 Usable Rules Processor

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

  half(0(x1))	→	0(s(s(half(x1))))	(7)
  half(s(s(x1)))	→	s(half(p(p(s(s(x1))))))	(8)
  p(s(x1))	→	x1	(2)
  p(0(x1))	→	0(s(s(p(x1))))	(1)

  1.1.1.1.1.1.1 Reduction Pair Processor

  Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals

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

  the pair

  f#(s(x1))

  →

  g#(s(x1))

  (14)

  could be deleted.

  1.1.1.1.1.1.1.1 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)))))	(21)
  half#(0(x1))	→	half#(x1)	(20)

  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
  [0(x1)]	=	1 · x1
  [half#(x1)]	=	1 · x1

  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)
  0(x1)	→	x1	(9)

  (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

  1.1.2.1 Monotonic Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

  [p(x1)]	=	1 · x1
  [s(x1)]	=	1 · x1
  [0(x1)]	=	2 + 2 · x1
  [half#(x1)]	=	1 · x1

  the pair

  half#(0(x1))

  →

  half#(x1)

  (20)

  and the rule

  0(x1)

  →

  x1

  (9)

  could be deleted.

  1.1.2.1.1 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.

  1.1.2.1.1.1 Usable Rules Processor

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

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

  1.1.2.1.1.1.1 Reduction Pair Processor

  Using the linear polynomial interpretation over the naturals

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

  the pair

  half#(s(s(x1)))

  →

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

  (21)

  could be deleted.

  1.1.2.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  p#(p(s(x1)))	→	p#(x1)	(13)
  p#(0(x1))	→	p#(x1)	(12)

  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
  [0(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.

  1.1.3.1 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)	(13)
  1	>	1
  p#(0(x1))	→	p#(x1)	(12)
  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)

  (30)

  1.1.4 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [0(x1)]	=	1 · x1
  [rd#(x1)]	=	1 · x1

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

  1.1.4.1 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)	(30)
  1	>	1

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