Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/double)

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 3 components.

• The 1^st component contains the pair

  permute#(false,x,b)	→	permute#(ack(x,x),p(x),c)	(21)
  permute#(y,x,c)	→	permute#(x,y,a)	(24)
  permute#(x,y,a)	→	permute#(isZero(x),x,b)	(19)

  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 Usable Rules Processor

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

  isZero(0)	→	true	(16)
  isZero(s(x))	→	false	(17)
  ack(0,x)	→	plus(x,s(0))	(8)
  ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(10)
  ack(s(x),0)	→	ack(x,s(0))	(9)
  p(0)	→	0	(6)
  p(s(x))	→	x	(7)
  plus(0,y)	→	y	(11)
  plus(s(x),y)	→	plus(x,s(y))	(12)
  plus(x,s(0))	→	s(x)	(14)
  plus(x,s(s(y)))	→	s(plus(s(x),y))	(13)
  plus(x,0)	→	x	(15)

  1.1.1.1.1 Innermost Lhss Removal Processor

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

  p(0)

  p(s(x0))

  ack(0,x0)

  ack(s(x0),0)

  ack(s(x0),s(x1))

  plus(0,x0)

  plus(s(x0),x1)

  plus(x0,s(s(x1)))

  plus(x0,s(0))

  plus(x0,0)

  isZero(0)

  isZero(s(x0))

  1.1.1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 1 to get the following set of pairs

  permute#(0,y1,a)	→	permute#(true,0,b)	(31)
  permute#(s(x0),y1,a)	→	permute#(false,s(x0),b)	(32)

  1.1.1.1.1.1.1 Dependency Graph Processor

  The dependency pairs are split into 1 component.

  • The 1^st component contains the pair

    permute#(y,x,c)	→	permute#(x,y,a)	(24)
    permute#(s(x0),y1,a)	→	permute#(false,s(x0),b)	(32)
    permute#(false,x,b)	→	permute#(ack(x,x),p(x),c)	(21)

    1.1.1.1.1.1.1.1 Usable Rules Processor

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

    ack(0,x)	→	plus(x,s(0))	(8)
    ack(s(x),0)	→	ack(x,s(0))	(9)
    ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(10)
    p(0)	→	0	(6)
    p(s(x))	→	x	(7)
    plus(0,y)	→	y	(11)
    plus(s(x),y)	→	plus(x,s(y))	(12)
    plus(x,s(0))	→	s(x)	(14)
    plus(x,s(s(y)))	→	s(plus(s(x),y))	(13)
    plus(x,0)	→	x	(15)

    1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

    p(0)

    p(s(x0))

    ack(0,x0)

    ack(s(x0),0)

    ack(s(x0),s(x1))

    plus(0,x0)

    plus(s(x0),x1)

    plus(x0,s(s(x1)))

    plus(x0,s(0))

    plus(x0,0)

    1.1.1.1.1.1.1.1.1.1 Instantiation Processor

    We instantiate the pair to the following set of pairs

    permute#(false,s(z0),b)

    →

    permute#(ack(s(z0),s(z0)),p(s(z0)),c)

    (33)

    1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

    ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(10)
    ack(s(x),0)	→	ack(x,s(0))	(9)
    p(s(x))	→	x	(7)
    ack(0,x)	→	plus(x,s(0))	(8)
    plus(0,y)	→	y	(11)
    plus(s(x),y)	→	plus(x,s(y))	(12)
    plus(x,s(0))	→	s(x)	(14)
    plus(x,s(s(y)))	→	s(plus(s(x),y))	(13)
    plus(x,0)	→	x	(15)

    1.1.1.1.1.1.1.1.1.1.1.1 Rewriting Processor

    We rewrite the right hand side of the pair resulting in

    →

    1.1.1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

    ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(10)
    ack(s(x),0)	→	ack(x,s(0))	(9)
    ack(0,x)	→	plus(x,s(0))	(8)
    plus(0,y)	→	y	(11)
    plus(s(x),y)	→	plus(x,s(y))	(12)
    plus(x,s(0))	→	s(x)	(14)
    plus(x,s(s(y)))	→	s(plus(s(x),y))	(13)
    plus(x,0)	→	x	(15)

    1.1.1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

    ack(0,x0)

    ack(s(x0),0)

    ack(s(x0),s(x1))

    plus(0,x0)

    plus(s(x0),x1)

    plus(x0,s(s(x1)))

    plus(x0,s(0))

    plus(x0,0)

    1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Narrowing Processor

    We consider all narrowings of the pair below position 1 to get the following set of pairs

    permute#(false,s(x0),b)

    →

    permute#(ack(x0,ack(s(x0),x0)),x0,c)

    (35)

    1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Forward Instantiation Processor

    We instantiate the pair to the following set of pairs

    permute#(x0,s(y_0),c)

    →

    permute#(s(y_0),x0,a)

    (36)

    1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Size-Change Termination

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

    permute#(false,s(x0),b)	→	permute#(ack(x0,ack(s(x0),x0)),x0,c)	(35)
    2	>	2
    permute#(x0,s(y_0),c)	→	permute#(s(y_0),x0,a)	(36)
    2	≥	1
    1	≥	2
    permute#(s(x0),y1,a)	→	permute#(false,s(x0),b)	(32)
    1	≥	2

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

• The 2^nd component contains the pair

  ack#(s(x),s(y))	→	ack#(x,ack(s(x),y))	(27)
  ack#(s(x),0)	→	ack#(x,s(0))	(26)
  ack#(s(x),s(y))	→	ack#(s(x),y)	(28)

  1.1.2 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 Usable Rules Processor

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

  ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(10)
  ack(s(x),0)	→	ack(x,s(0))	(9)
  ack(0,x)	→	plus(x,s(0))	(8)
  plus(0,y)	→	y	(11)
  plus(s(x),y)	→	plus(x,s(y))	(12)
  plus(x,s(0))	→	s(x)	(14)
  plus(x,s(s(y)))	→	s(plus(s(x),y))	(13)
  plus(x,0)	→	x	(15)

  1.1.2.1.1 Innermost Lhss Removal Processor

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

  ack(0,x0)

  ack(s(x0),0)

  ack(s(x0),s(x1))

  plus(0,x0)

  plus(s(x0),x1)

  plus(x0,s(s(x1)))

  plus(x0,s(0))

  plus(x0,0)

  1.1.2.1.1.1 Size-Change Termination

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

  ack#(s(x),0)	→	ack#(x,s(0))	(26)
  1	>	1
  ack#(s(x),s(y))	→	ack#(x,ack(s(x),y))	(27)
  1	>	1
  ack#(s(x),s(y))	→	ack#(s(x),y)	(28)
  1	≥	1
  2	>	2

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

• The 3^rd component contains the pair

  plus#(x,s(s(y)))	→	plus#(s(x),y)	(30)
  plus#(s(x),y)	→	plus#(x,s(y))	(29)

  1.1.3 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 Usable Rules Processor

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

  There are no rules.

  1.1.3.1.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.3.1.1.1 Monotonic Reduction Pair Processor

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

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

  the pairs

  plus#(x,s(s(y)))	→	plus#(s(x),y)	(30)
  plus#(s(x),y)	→	plus#(x,s(y))	(29)

  and the rules

  double(x)	→	permute(x,x,a)	(1)
  permute(x,y,a)	→	permute(isZero(x),x,b)	(2)
  permute(false,x,b)	→	permute(ack(x,x),p(x),c)	(3)
  permute(true,x,b)	→	0	(4)
  permute(y,x,c)	→	s(s(permute(x,y,a)))	(5)
  p(0)	→	0	(6)
  p(s(x))	→	x	(7)
  ack(0,x)	→	plus(x,s(0))	(8)
  ack(s(x),0)	→	ack(x,s(0))	(9)
  ack(s(x),s(y))	→	ack(x,ack(s(x),y))	(10)
  plus(0,y)	→	y	(11)
  plus(s(x),y)	→	plus(x,s(y))	(12)
  plus(x,s(s(y)))	→	s(plus(s(x),y))	(13)
  plus(x,s(0))	→	s(x)	(14)
  plus(x,0)	→	x	(15)
  isZero(0)	→	true	(16)
  isZero(s(x))	→	false	(17)

  could be deleted.

  1.1.3.1.1.1.1 P is empty

  There are no pairs anymore.