Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/aprove2)

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

  div#(x,y)	→	ify#(ge(y,s(0)),x,y)	(13)
  ify#(true,x,y)	→	if#(ge(x,y),x,y)	(15)
  if#(true,x,y)	→	div#(minus(x,y),y)	(17)

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

  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(s(x0),s(x1))

  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)	→	ify#(false,y0,0)	(19)
  div#(y0,s(x0))	→	ify#(ge(x0,0),y0,s(x0))	(20)

  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,x,y)	→	div#(minus(x,y),y)	(17)
    div#(y0,s(x0))	→	ify#(ge(x0,0),y0,s(x0))	(20)
    ify#(true,x,y)	→	if#(ge(x,y),x,y)	(15)

    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 Narrowing Processor

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

    ify#(true,x0,0)	→	if#(true,x0,0)	(22)
    ify#(true,0,s(x0))	→	if#(false,0,s(x0))	(23)
    ify#(true,s(x0),s(x1))	→	if#(ge(x0,x1),s(x0),s(x1))	(24)

    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#(y0,s(x0))	→	ify#(true,y0,s(x0))	(21)
      ify#(true,s(x0),s(x1))	→	if#(ge(x0,x1),s(x0),s(x1))	(24)
      if#(true,x,y)	→	div#(minus(x,y),y)	(17)

      1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      if#(true,s(z0),s(z1))

      →

      div#(minus(s(z0),s(z1)),s(z1))

      (25)

      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 Forward Instantiation Processor

      We instantiate the pair to the following set of pairs

      div#(s(y_0),s(x1))

      →

      ify#(true,s(y_0),s(x1))

      (27)

      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,s(x0),s(0))	→	div#(x0,s(0))	(28)
      if#(true,s(s(x0)),s(s(x1)))	→	div#(minus(x0,x1),s(s(x1)))	(29)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      div#(s(x0),s(0))	→	ify#(true,s(x0),s(0))	(30)
      div#(s(x0),s(s(z1)))	→	ify#(true,s(x0),s(s(z1)))	(31)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Instantiation Processor

      We instantiate the pair to the following set of pairs

      ify#(true,s(z0),s(0))	→	if#(ge(z0,0),s(z0),s(0))	(32)
      ify#(true,s(z0),s(s(z1)))	→	if#(ge(z0,s(z1)),s(z0),s(s(z1)))	(33)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor

      The dependency pairs are split into 2 components.

      • The 1^st component contains the pair

        div#(s(x0),s(0))	→	ify#(true,s(x0),s(0))	(30)
        ify#(true,s(z0),s(0))	→	if#(ge(z0,0),s(z0),s(0))	(32)
        if#(true,s(x0),s(0))	→	div#(x0,s(0))	(28)

        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(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 Rewriting Processor

        We rewrite the right hand side of the pair resulting in

        →

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Usable Rules Processor

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

        There are no rules.

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

        There are no lhss.

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Forward Instantiation Processor

        We instantiate the pair to the following set of pairs

        if#(true,s(s(y_0)),s(0))

        →

        div#(s(y_0),s(0))

        (35)

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Forward Instantiation Processor

        We instantiate the pair to the following set of pairs

        ify#(true,s(s(y_0)),s(0))

        →

        if#(true,s(s(y_0)),s(0))

        (36)

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Size-Change Termination

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

        ify#(true,s(s(y_0)),s(0))	→	if#(true,s(s(y_0)),s(0))	(36)
        1	≥	1
        2	≥	2
        3	≥	3
        if#(true,s(s(y_0)),s(0))	→	div#(s(y_0),s(0))	(35)
        2	>	1
        3	≥	2
        div#(s(x0),s(0))	→	ify#(true,s(x0),s(0))	(30)
        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

        div#(s(x0),s(s(z1)))	→	ify#(true,s(x0),s(s(z1)))	(31)
        ify#(true,s(z0),s(s(z1)))	→	if#(ge(z0,s(z1)),s(z0),s(s(z1)))	(33)
        if#(true,s(s(x0)),s(s(x1)))	→	div#(minus(x0,x1),s(s(x1)))	(29)

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.2 Forward Instantiation Processor

        We instantiate the pair to the following set of pairs

        ify#(true,s(s(y_1)),s(s(x1)))

        →

        if#(ge(s(y_1),s(x1)),s(s(y_1)),s(s(x1)))

        (37)

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.2.1 Rewriting Processor

        We rewrite the right hand side of the pair resulting in

        →

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.2.1.1 Reduction Pair Processor with Usable Rules

        Using the linear polynomial interpretation over the naturals

        [div#(x1, x2)]	=	2 + 2 · x1 + 2 · x2
        [minus(x1, x2)]	=	2 + x1
        [0]	=	2
        [s(x1)]	=	2 + 2 · x1
        [if#(x1, x2, x3)]	=	-2 + 2 · x2 + 2 · x3
        [ge(x1, x2)]	=	2 · x1
        [true]	=	0
        [false]	=	1
        [ify#(x1, x2, x3)]	=	-1 + 2 · x1 + 2 · x2 + 2 · x3

        together with the usable rules

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

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

        div#(s(x0),s(s(z1)))	→	ify#(true,s(x0),s(s(z1)))	(31)
        if#(true,s(s(x0)),s(s(x1)))	→	div#(minus(x0,x1),s(s(x1)))	(29)
        ify#(true,s(s(y_1)),s(s(x1)))	→	if#(ge(y_1,x1),s(s(y_1)),s(s(x1)))	(38)

        could be deleted.

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.2.1.1.1 P is empty

        There are no pairs anymore.

• The 2^nd component contains the pair

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

  →

  ge#(x,y)

  (11)

  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.

  ge#(s(x),s(y))	→	ge#(x,y)	(11)
  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)

  (12)

  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.

  minus#(s(x),s(y))	→	minus#(x,y)	(12)
  1	>	1
  2	>	2

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