Certification Problem

Input (TPDB TRS_Standard/AProVE_07/thiemann01)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

Hence, it suffices to show innermost termination in the following.

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

  if#(false,x,s(y),z)	→	quot#(minus(x,s(y)),s(y),z)	(20)
  quot#(x,y,z)	→	if#(zero(x),x,y,plus(z,s(0)))	(16)

  1.1.1.1 Usable Rules Processor

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

  zero(s(x))	→	false	(6)
  zero(0)	→	true	(7)
  plus(0,y)	→	y	(4)
  plus(s(x),y)	→	plus(x,s(y))	(5)
  minus(0,y)	→	0	(1)
  minus(s(x),s(y))	→	minus(x,y)	(3)
  minus(x,0)	→	x	(2)

  1.1.1.1.1 Innermost Lhss Removal Processor

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

  minus(0,x0)

  minus(x0,0)

  minus(s(x0),s(x1))

  plus(0,x0)

  plus(s(x0),x1)

  zero(s(x0))

  zero(0)

  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

  quot#(s(x0),y1,y2)	→	if#(false,s(x0),y1,plus(y2,s(0)))	(22)
  quot#(0,y1,y2)	→	if#(true,0,y1,plus(y2,s(0)))	(23)

  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

    quot#(s(x0),y1,y2)	→	if#(false,s(x0),y1,plus(y2,s(0)))	(22)
    if#(false,x,s(y),z)	→	quot#(minus(x,s(y)),s(y),z)	(20)

    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.

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

    1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

    minus(0,x0)

    minus(x0,0)

    minus(s(x0),s(x1))

    plus(0,x0)

    plus(s(x0),x1)

    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

    if#(false,0,s(y1),y2)	→	quot#(0,s(y1),y2)	(24)
    if#(false,s(x0),s(x1),y2)	→	quot#(minus(x0,x1),s(x1),y2)	(25)

    1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      if#(false,s(x0),s(x1),y2)	→	quot#(minus(x0,x1),s(x1),y2)	(25)
      quot#(s(x0),y1,y2)	→	if#(false,s(x0),y1,plus(y2,s(0)))	(22)

      1.1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      quot#(s(x0),s(z1),z2)

      →

      if#(false,s(x0),s(z1),plus(z2,s(0)))

      (26)

      1.1.1.1.1.1.1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

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

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

      in combination with the following argument filter

      π(if#)	=	2
      π(s)	=	[1]
      π(quot#)	=	1
      π(minus)	=	1
      π(0)	=	[]

      together with the usable rules

      minus(0,y)	→	0	(1)
      minus(s(x),s(y))	→	minus(x,y)	(3)
      minus(x,0)	→	x	(2)

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

      if#(false,s(x0),s(x1),y2)

      →

      quot#(minus(x0,x1),s(x1),y2)

      (25)

      could be deleted.

      1.1.1.1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

      plus(0,y)	→	y	(4)
      plus(s(x),y)	→	plus(x,s(y))	(5)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

      plus(0,x0)

      plus(s(x0),x1)

      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.

      quot#(s(x0),s(z1),z2)	→	if#(false,s(x0),s(z1),plus(z2,s(0)))	(26)
      1	≥	2
      2	≥	3

      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#(x,y)

  (13)

  1.1.1.2 Usable Rules Processor

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

  There are no rules.

  1.1.1.2.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.1.2.1.1 Size-Change Termination

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

  minus#(s(x),s(y))	→	minus#(x,y)	(13)
  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#(s(x),y)

  →

  plus#(x,s(y))

  (14)

  1.1.1.3 Usable Rules Processor

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

  There are no rules.

  1.1.1.3.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.1.3.1.1 Size-Change Termination

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

  plus#(s(x),y)	→	plus#(x,s(y))	(14)
  1	>	1

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