Certification Problem

Input (TPDB TRS_Standard/AProVE_06/tower_sizeChange)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

1.1 Rule Removal

1.1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

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

1.1.1.1 Dependency Pair Transformation

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.

• The 1^st component contains the pair

  f#(b,y,x)	→	f#(a,half(x),exp(y))	(13)
  f#(a,s(x),y)	→	f#(b,y,s(x))	(12)

  1.1.1.1.1.1 Usable Rules Processor

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

  half(0)	→	double(0)	(9)
  half(s(0))	→	half(0)	(10)
  half(s(s(x)))	→	s(half(x))	(11)
  exp(0)	→	s(0)	(5)
  exp(s(x))	→	double(exp(x))	(6)
  double(0)	→	0	(7)
  double(s(x))	→	s(s(double(x)))	(8)

  1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

  exp(0)

  exp(s(x0))

  double(0)

  double(s(x0))

  half(0)

  half(s(0))

  half(s(s(x0)))

  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

  f#(b,y0,0)	→	f#(a,double(0),exp(y0))	(22)
  f#(b,y0,s(0))	→	f#(a,half(0),exp(y0))	(23)
  f#(b,y0,s(s(x0)))	→	f#(a,s(half(x0)),exp(y0))	(24)

  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

    f#(b,y0,s(0))	→	f#(a,half(0),exp(y0))	(23)
    f#(a,s(x),y)	→	f#(b,y,s(x))	(12)
    f#(b,y0,s(s(x0)))	→	f#(a,s(half(x0)),exp(y0))	(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 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 Dependency Graph Processor

    The dependency pairs are split into 1 component.

    • The 1^st component contains the pair

      f#(b,y0,s(s(x0)))	→	f#(a,s(half(x0)),exp(y0))	(24)
      f#(a,s(x),y)	→	f#(b,y,s(x))	(12)

      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

      f#(a,s(s(y_1)),x1)

      →

      f#(b,x1,s(s(y_1)))

      (27)

      1.1.1.1.1.1.1.1.1.1.1.1.1.1 Semantic Labeling Processor

      The following interpretations form a model of the rules.

      As carrier we take the set {0,1}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 2):

      [s(x1)]	=	0
      [exp(x1)]	=	0
      [a]	=	1
      [b]	=	0
      [half(x1)]	=	0
      [0]	=	0
      [f#(x1, x2, x3)]	=	0
      [double(x1)]	=	0

      We obtain the set of labeled pairs

      f#000(b,y0,s0(s0(x0)))	→	f#100(a,s0(half0(x0)),exp0(y0))	(28)
      f#100(a,s0(s0(y_1)),x1)	→	f#000(b,x1,s0(s0(y_1)))	(29)
      f#101(a,s0(s0(y_1)),x1)	→	f#010(b,x1,s0(s0(y_1)))	(30)
      f#100(a,s0(s1(y_1)),x1)	→	f#000(b,x1,s0(s1(y_1)))	(31)
      f#101(a,s0(s1(y_1)),x1)	→	f#010(b,x1,s0(s1(y_1)))	(32)
      f#000(b,y0,s0(s1(x0)))	→	f#100(a,s0(half1(x0)),exp0(y0))	(33)
      f#010(b,y0,s0(s0(x0)))	→	f#100(a,s0(half0(x0)),exp1(y0))	(34)
      f#010(b,y0,s0(s1(x0)))	→	f#100(a,s0(half1(x0)),exp1(y0))	(35)

      and the set of labeled rules:

      half0(0)	→	double0(0)	(36)
      half0(s0(0))	→	half0(0)	(37)
      half0(s0(s0(x)))	→	s0(half0(x))	(38)
      half0(s0(s1(x)))	→	s0(half1(x))	(39)
      exp0(0)	→	s0(0)	(40)
      exp0(s0(x))	→	double0(exp0(x))	(41)
      exp0(s1(x))	→	double0(exp1(x))	(42)
      double0(0)	→	0	(43)
      double0(s0(x))	→	s0(s0(double0(x)))	(44)
      double0(s1(x))	→	s0(s0(double1(x)))	(45)

      Innermost rewriting w.r.t. the following left-hand sides is considered:

      exp0(0)

      exp0(s0(x0))

      exp0(s1(x0))

      double0(0)

      double0(s0(x0))

      double0(s1(x0))

      half0(0)

      half0(s0(0))

      half0(s0(s0(x0)))

      half0(s0(s1(x0)))

      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

        f#100(a,s0(s0(y_1)),x1)	→	f#000(b,x1,s0(s0(y_1)))	(29)
        f#000(b,y0,s0(s0(x0)))	→	f#100(a,s0(half0(x0)),exp0(y0))	(28)

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

        Using the linear polynomial interpretation over the naturals

        [half0(x1)]	=	1 · x1
        [0]	=	0
        [double0(x1)]	=	1 · x1
        [s0(x1)]	=	1 · x1
        [s1(x1)]	=	1 + 1 · x1
        [half1(x1)]	=	1 + 1 · x1
        [exp0(x1)]	=	1 · x1
        [exp1(x1)]	=	1 + 1 · x1
        [double1(x1)]	=	1 + 1 · x1
        [f#100(x1, x2, x3)]	=	1 · x1 + 1 · x2 + 1 · x3
        [a]	=	1
        [f#000(x1, x2, x3)]	=	1 + 1 · x1 + 1 · x2 + 1 · x3
        [b]	=	0

        together with the usable rules

        half0(0)	→	double0(0)	(36)
        half0(s0(0))	→	half0(0)	(37)
        half0(s0(s0(x)))	→	s0(half0(x))	(38)
        half0(s0(s1(x)))	→	s0(half1(x))	(39)
        exp0(0)	→	s0(0)	(40)
        exp0(s0(x))	→	double0(exp0(x))	(41)
        exp0(s1(x))	→	double0(exp1(x))	(42)
        double0(0)	→	0	(43)
        double0(s0(x))	→	s0(s0(double0(x)))	(44)
        double0(s1(x))	→	s0(s0(double1(x)))	(45)

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

        half0(s0(s1(x)))	→	s0(half1(x))	(39)
        exp0(s1(x))	→	double0(exp1(x))	(42)
        double0(s1(x))	→	s0(s0(double1(x)))	(45)

        could be deleted.

        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 linear polynomial interpretation over the naturals

        [f#100(x1, x2, x3)]	=	1 · x1 + 1 · x2
        [a]	=	0
        [s0(x1)]	=	1 + 1 · x1
        [f#000(x1, x2, x3)]	=	1 · x1 + 1 · x3
        [b]	=	0
        [half0(x1)]	=	1 · x1
        [exp0(x1)]	=	0
        [0]	=	0
        [double0(x1)]	=	0

        together with the usable rules

        half0(0)	→	double0(0)	(36)
        half0(s0(0))	→	half0(0)	(37)
        half0(s0(s0(x)))	→	s0(half0(x))	(38)
        double0(0)	→	0	(43)

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

        f#000(b,y0,s0(s0(x0)))

        →

        f#100(a,s0(half0(x0)),exp0(y0))

        (28)

        could be deleted.

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

        Using the linear polynomial interpretation over the naturals

        [f#100(x1, x2, x3)]	=	1 + 1 · x1 + 1 · x2 + 1 · x3
        [a]	=	1
        [s0(x1)]	=	1 · x1
        [f#000(x1, x2, x3)]	=	1 + 1 · x1 + 1 · x2 + 1 · x3
        [b]	=	1

        having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair

        f#100(a,s0(s0(y_1)),x1)

        →

        f#000(b,x1,s0(s0(y_1)))

        (29)

        and no rules could be deleted.

        1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 P is empty

        There are no pairs anymore.

• The 2^nd component contains the pair

  half#(s(s(x)))

  →

  half#(x)

  (21)

  1.1.1.1.1.2 Usable Rules Processor

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

  There are no rules.

  1.1.1.1.1.2.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.1.1.1.2.1.1 Size-Change Termination

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

  half#(s(s(x)))	→	half#(x)	(21)
  1	>	1

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

• The 3^rd component contains the pair

  exp#(s(x))

  →

  exp#(x)

  (17)

  1.1.1.1.1.3 Usable Rules Processor

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

  There are no rules.

  1.1.1.1.1.3.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.1.1.1.3.1.1 Size-Change Termination

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

  exp#(s(x))	→	exp#(x)	(17)
  1	>	1

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

• The 4^th component contains the pair

  double#(s(x))

  →

  double#(x)

  (18)

  1.1.1.1.1.4 Usable Rules Processor

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

  There are no rules.

  1.1.1.1.1.4.1 Innermost Lhss Removal Processor

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

  There are no lhss.

  1.1.1.1.1.4.1.1 Size-Change Termination

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

  double#(s(x))	→	double#(x)	(18)
  1	>	1

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