Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/aprove5)

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

  div#(plus(x,y),z)	→	div#(x,z)	(20)
  div#(s(x),s(y))	→	div#(minus(x,y),s(y))	(17)
  div#(plus(x,y),z)	→	div#(y,z)	(21)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

  prec(plus)	=	1		weight(plus)	=	1
  prec(0)	=	2		weight(0)	=	2
  prec(p)	=	0		weight(p)	=	1

  in combination with the following argument filter

  π(div#)	=	1
  π(plus)	=	[1,2]
  π(s)	=	1
  π(minus)	=	1
  π(0)	=	[]
  π(p)	=	[]

  together with the usable rules

  minus(x,0)	→	x	(1)
  minus(0,y)	→	0	(2)
  minus(x,plus(y,z))	→	minus(minus(x,y),z)	(4)
  minus(s(x),s(y))	→	minus(p(s(x)),p(s(y)))	(3)
  p(s(s(x)))	→	s(p(s(x)))	(5)

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

  div#(plus(x,y),z)	→	div#(x,z)	(20)
  div#(plus(x,y),z)	→	div#(y,z)	(21)

  could be deleted.

  1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

  prec(s)	=	0		weight(s)	=	1
  prec(0)	=	1		weight(0)	=	1

  in combination with the following argument filter

  π(div#)	=	1
  π(s)	=	[1]
  π(minus)	=	1
  π(0)	=	[]
  π(p)	=	1

  together with the usable rules

  minus(x,0)	→	x	(1)
  minus(0,y)	→	0	(2)
  minus(x,plus(y,z))	→	minus(minus(x,y),z)	(4)
  minus(s(x),s(y))	→	minus(p(s(x)),p(s(y)))	(3)
  p(s(s(x)))	→	s(p(s(x)))	(5)

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

  div#(s(x),s(y))

  →

  div#(minus(x,y),s(y))

  (17)

  could be deleted.

  1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  plus#(s(x),y)

  →

  plus#(y,minus(s(x),s(0)))

  (22)

  1.1.2 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [minus(x1, x2)]	=	1 · x1 + 1 · x2
  [plus(x1, x2)]	=	1 · x1 + 1 · x2
  [s(x1)]	=	1 · x1
  [p(x1)]	=	1 · x1
  [0]	=	0
  [plus#(x1, x2)]	=	1 · x1 + 1 · x2

  together with the usable rules

  minus(x,plus(y,z))	→	minus(minus(x,y),z)	(4)
  minus(s(x),s(y))	→	minus(p(s(x)),p(s(y)))	(3)
  minus(x,0)	→	x	(1)
  minus(0,y)	→	0	(2)
  p(s(s(x)))	→	s(p(s(x)))	(5)

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

  1.1.2.1 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 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [minus(x1, x2)]	=	1 · x1 + 1 · x2
  [s(x1)]	=	1 · x1
  [p(x1)]	=	1 · x1
  [plus(x1, x2)]	=	2 + 2 · x1 + 2 · x2
  [0]	=	0
  [plus#(x1, x2)]	=	1 · x1 + 1 · x2

  together with the usable rules

  minus(s(x),s(y))	→	minus(p(s(x)),p(s(y)))	(3)
  minus(x,plus(y,z))	→	minus(minus(x,y),z)	(4)
  p(s(s(x)))	→	s(p(s(x)))	(5)
  minus(x,0)	→	x	(1)
  minus(0,y)	→	0	(2)

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

  minus(x,plus(y,z))

  →

  minus(minus(x,y),z)

  (4)

  could be deleted.

  1.1.2.1.1.1 Rewriting Processor

  We rewrite the right hand side of the pair resulting in

  →

  1.1.2.1.1.1.1 Usable Rules Processor

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

  p(s(s(x)))	→	s(p(s(x)))	(5)
  minus(0,y)	→	0	(2)

  1.1.2.1.1.1.1.1 Reduction Pair Processor with Usable Rules

  Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions

  prec(plus#)	=	1		weight(plus#)	=	1
  prec(s)	=	0		weight(s)	=	3
  prec(minus)	=	3		weight(minus)	=	2
  prec(0)	=	2		weight(0)	=	1

  in combination with the following argument filter

  π(plus#)	=	[1,2]
  π(s)	=	[]
  π(minus)	=	[]
  π(0)	=	[]

  together with the usable rule

  minus(0,y)

  →

  0

  (2)

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

  plus#(s(x),y)

  →

  plus#(y,minus(p(s(x)),p(s(0))))

  (24)

  could be deleted.

  1.1.2.1.1.1.1.1.1 P is empty

  There are no pairs anymore.

• The 3^rd component contains the pair

  minus#(x,plus(y,z))	→	minus#(minus(x,y),z)	(14)
  minus#(s(x),s(y))	→	minus#(p(s(x)),p(s(y)))	(11)
  minus#(x,plus(y,z))	→	minus#(x,y)	(15)

  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
  [minus(x1, x2)]	=	1 · x1 + 1 · x2
  [0]	=	0
  [plus(x1, x2)]	=	1 · x1 + 1 · x2
  [minus#(x1, x2)]	=	1 · x1 + 1 · x2

  together with the usable rules

  p(s(s(x)))	→	s(p(s(x)))	(5)
  minus(x,0)	→	x	(1)
  minus(0,y)	→	0	(2)
  minus(x,plus(y,z))	→	minus(minus(x,y),z)	(4)
  minus(s(x),s(y))	→	minus(p(s(x)),p(s(y)))	(3)

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

  1.1.3.1 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.

  1.1.3.1.1 Dependency Graph Processor

  The dependency pairs are split into 2 components.

  • The 1^st component contains the pair

    minus#(x,plus(y,z))	→	minus#(x,y)	(15)
    minus#(x,plus(y,z))	→	minus#(minus(x,y),z)	(14)

    1.1.3.1.1.1 Size-Change Termination

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

    minus#(x,plus(y,z))	→	minus#(x,y)	(15)
    1	≥	1
    2	>	2
    minus#(x,plus(y,z))	→	minus#(minus(x,y),z)	(14)
    2	>	2

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

  • The 2^nd component contains the pair

    minus#(s(x),s(y))

    →

    minus#(p(s(x)),p(s(y)))

    (11)

    1.1.3.1.1.2 Usable Rules Processor

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

    p(s(s(x)))

    →

    s(p(s(x)))

    (5)

    1.1.3.1.1.2.1 Innermost Lhss Removal Processor

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

    p(s(s(x0)))

    1.1.3.1.1.2.1.1 Reduction Pair Processor

    Using the linear polynomial interpretation over the naturals

    [minus#(x1, x2)]	=	2 + 2 · x1 + 2 · x2
    [p(x1)]	=	-2 + x1
    [s(x1)]	=	1 + 2 · x1

    the pair

    minus#(s(x),s(y))

    →

    minus#(p(s(x)),p(s(y)))

    (11)

    could be deleted.

    1.1.3.1.1.2.1.1.1 P is empty

    There are no pairs anymore.

• The 4^th component contains the pair

  p#(s(s(x)))

  →

  p#(s(x))

  (16)

  1.1.4 Monotonic Reduction Pair Processor with Usable Rules

  Using the linear polynomial interpretation over the naturals

  [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.

  1.1.4.1 Size-Change Termination

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

  p#(s(s(x)))	→	p#(s(x))	(16)
  1	>	1

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