Certification Problem

Input (TPDB TRS_Innermost/Mixed_innermost/thiemann26i)

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

  div#(x,y,z)	→	if#(ge(y,s(0)),ge(x,y),x,y,z)	(17)
  if#(true,true,x,y,z)	→	div#(minus(x,y),y,id_inc(z))	(20)

  1.1.1 Usable Rules Processor

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

  minus(x,0)	→	x	(4)
  minus(0,y)	→	0	(5)
  minus(s(x),s(y))	→	minus(x,y)	(6)
  id_inc(x)	→	x	(7)
  id_inc(x)	→	s(x)	(8)
  ge(0,s(y))	→	false	(2)
  ge(s(x),s(y))	→	ge(x,y)	(3)
  ge(x,0)	→	true	(1)

  1.1.1.1 Innermost Lhss Removal Processor

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

  ge(x0,0)

  ge(0,s(x0))

  ge(s(x0),s(x1))

  minus(x0,0)

  minus(0,x0)

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

  id_inc(x0)

  1.1.1.1.1 Narrowing Processor

  We consider all narrowings of the pair below position 3 to get the following set of pairs

  if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,x0)	(23)
  if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,s(x0))	(24)

  1.1.1.1.1.1 Usable Rules Processor

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

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

  1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

  ge(x0,0)

  ge(0,s(x0))

  ge(s(x0),s(x1))

  minus(x0,0)

  minus(0,x0)

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

  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

  div#(y0,0,y2)	→	if#(false,ge(y0,0),y0,0,y2)	(25)
  div#(y0,s(x0),y2)	→	if#(ge(x0,0),ge(y0,s(x0)),y0,s(x0),y2)	(26)

  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

    div#(y0,s(x0),y2)	→	if#(ge(x0,0),ge(y0,s(x0)),y0,s(x0),y2)	(26)
    if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,x0)	(23)
    if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,s(x0))	(24)

    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 Narrowing Processor

    We consider all narrowings of the pair below position 2 to get the following set of pairs

    div#(0,s(x0),y2)	→	if#(true,false,0,s(x0),y2)	(28)
    div#(s(x0),s(x1),y2)	→	if#(true,ge(x0,x1),s(x0),s(x1),y2)	(29)

    1.1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      div#(s(x0),s(x1),y2)	→	if#(true,ge(x0,x1),s(x0),s(x1),y2)	(29)
      if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,x0)	(23)
      if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,s(x0))	(24)

      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

      if#(true,true,x0,0,y2)	→	div#(x0,0,y2)	(30)
      if#(true,true,0,x0,y2)	→	div#(0,x0,y2)	(31)
      if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),y2)	(32)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 1 component.

      • The 1^st component contains the pair

        if#(true,true,y0,y1,x0)	→	div#(minus(y0,y1),y1,s(x0))	(24)
        div#(s(x0),s(x1),y2)	→	if#(true,ge(x0,x1),s(x0),s(x1),y2)	(29)
        if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),y2)	(32)

        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

        if#(true,true,x0,0,y2)	→	div#(x0,0,s(y2))	(33)
        if#(true,true,0,x0,y2)	→	div#(0,x0,s(y2))	(34)
        if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),s(y2))	(35)

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

        The dependency pairs are split into 1 component.

        • The 1^st component contains the pair

          if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),y2)	(32)
          div#(s(x0),s(x1),y2)	→	if#(true,ge(x0,x1),s(x0),s(x1),y2)	(29)
          if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),s(y2))	(35)

          1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 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#)	=	3
          π(s)	=	[1]
          π(div#)	=	1
          π(minus)	=	1
          π(0)	=	[]

          together with the usable rules

          minus(x,0)	→	x	(4)
          minus(0,y)	→	0	(5)
          minus(s(x),s(y))	→	minus(x,y)	(6)

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

          if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),y2)	(32)
          if#(true,true,s(x0),s(x1),y2)	→	div#(minus(x0,x1),s(x1),s(y2))	(35)

          could be deleted.

          1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

          ge(0,s(y))	→	false	(2)
          ge(s(x),s(y))	→	ge(x,y)	(3)
          ge(x,0)	→	true	(1)

          1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

          ge(x0,0)

          ge(0,s(x0))

          ge(s(x0),s(x1))

          1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Size-Change Termination

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

          div#(s(x0),s(x1),y2)	→	if#(true,ge(x0,x1),s(x0),s(x1),y2)	(29)
          1	≥	3
          2	≥	4
          3	≥	5

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

• The 2^nd component contains the pair

  ge#(s(x),s(y))

  →

  ge#(x,y)

  (14)

  1.1.2 Usable Rules Processor

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

  There are no rules.

  1.1.2.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  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.

  ge#(s(x),s(y))	→	ge#(x,y)	(14)
  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

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

  →

  minus#(x,y)

  (15)

  1.1.3 Usable Rules Processor

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

  There are no rules.

  1.1.3.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.3.1.1 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)	(15)
  1	>	1
  2	>	2

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