Certification Problem

Input (TPDB TRS_Standard/AProVE_07/thiemann19)

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#(true,x)	→	1024_1#(s(x))	(16)
  1024_1#(x)	→	if#(lt(x,10),x)	(12)

  1.1.1.1 Usable Rules Processor

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

  10	→	double(s(double(s(s(0)))))	(10)
  lt(0,s(y))	→	true	(5)
  lt(x,0)	→	false	(6)
  lt(s(x),s(y))	→	lt(x,y)	(7)
  double(s(x))	→	s(s(double(x)))	(9)
  double(0)	→	0	(8)

  1.1.1.1.1 Innermost Lhss Removal Processor

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

  lt(0,s(x0))

  lt(x0,0)

  lt(s(x0),s(x1))

  double(0)

  double(s(x0))

  10

  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 Usable Rules Processor

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

  double(s(x))	→	s(s(double(x)))	(9)
  lt(0,s(y))	→	true	(5)
  lt(x,0)	→	false	(6)
  lt(s(x),s(y))	→	lt(x,y)	(7)
  double(0)	→	0	(8)

  1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

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

  lt(0,s(x0))

  lt(x0,0)

  lt(s(x0),s(x1))

  double(0)

  double(s(x0))

  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 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 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 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 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 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 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 Usable Rules Processor

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

  double(0)	→	0	(8)
  lt(0,s(y))	→	true	(5)
  lt(s(x),s(y))	→	lt(x,y)	(7)
  lt(x,0)	→	false	(6)

  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 Usable Rules Processor

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

  lt(0,s(y))	→	true	(5)
  lt(s(x),s(y))	→	lt(x,y)	(7)
  lt(x,0)	→	false	(6)

  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.

  lt(0,s(x0))

  lt(x0,0)

  lt(s(x0),s(x1))

  1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Complex Constant Removal Processor

  We replace the term s(s(s(s(s(s(s(s(s(s(0)))))))))) by a fresh variable. This results in the following new pairs.

  	→
  	→

  1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1.1 Reduction Pair Processor

  We apply the generic reduction pair processor using the linear polynomial interpretation over the naturals

  [c]	=	-2
  [1024_1^1(x1, x2)]	=	-1 · x1 + 1 · x2 + -1
  [IF(x1, x2, x3)]	=	-1 · x1 + -1 · x2 + 1 · x3 + -1
  [true]	=	0
  [s(x1)]	=	1 · x1 + 1
  [lt(x1, x2)]	=	0
  [0]	=	0
  [false]	=	0

  The pair(s)

  IF(true,x,x_removed)

  →

  1024_1^1(s(x),x_removed)

  (31)

  are strictly oriented and the pair(s)

  IF(true,x,x_removed)

  →

  1024_1^1(s(x),x_removed)

  (31)

  are bounded w.r.t. the constant c.

  The following constraints are generated for the pairs.

  • IF(true,0,s(x12))≥c
  • (IF(true,x14,x13)≥c⟹IF(true,s(x14),s(x13))≥c)
  • IF(true,0,s(x12)) > 1024_1^1(s(0),s(x12))
  • (IF(true,x14,x13) > 1024_1^1(s(x14),x13)⟹IF(true,s(x14),s(x13)) > 1024_1^1(s(s(x14)),s(x13)))
  • 1024_1^1(s(x6),x7)≥IF(lt(s(x6),x7),s(x6),x7)

  The details are shown below:

  • For the chain we build the initial constraint

    (IF(lt(x2,x3),x2,x3)→^*IF(true,x4,x5)⟹IF(true,x4,x5)≥c)

    which is simplified as follows.

    • Applying Rule "Same Constructor" results in
      (lt(x2,x3)→^*true⟹x2→^*x4⟹x3→^*x5⟹IF(true,x4,x5)≥c)
    • Applying Rule "Variable in Equation" allows to substitute x4 by x2 which results in
      (lt(x2,x3)→^*true⟹x3→^*x5⟹IF(true,x2,x5)≥c)
    • Applying Rule "Variable in Equation" allows to substitute x5 by x3 which results in
      (lt(x2,x3)→^*true⟹IF(true,x2,x3)≥c)
    • Applying Rule "Induction" on lt(x2,x3)→^*true results in the following 3 new constraints.
      1. (true→^*true⟹IF(true,0,s(x12))≥c)
         • Applying Rule "Same Constructor" results in
           IF(true,0,s(x12))≥c

         • This constraint is kept as final constraint.

      2. (lt(x14,x13)→^*true⟹(lt(x14,x13)→^*true⟹IF(true,x14,x13)≥c)⟹IF(true,s(x14),s(x13))≥c)
         • Applying Rule "Simplify Conditions" results in
           (IF(true,x14,x13)≥c⟹IF(true,s(x14),s(x13))≥c)

         • This constraint is kept as final constraint.

      3. (false→^*true⟹IF(true,x15,0)≥c)
         • Applying Rule "Different Constructors" allows to drop this constraint.
  • For the chain we build the initial constraint

    (IF(lt(x2,x3),x2,x3)→^*IF(true,x4,x5)⟹IF(true,x4,x5) > 1024_1^1(s(x4),x5))

    which is simplified as follows.

    • Applying Rule "Same Constructor" results in
      (lt(x2,x3)→^*true⟹x2→^*x4⟹x3→^*x5⟹IF(true,x4,x5) > 1024_1^1(s(x4),x5))
    • Applying Rule "Variable in Equation" allows to substitute x4 by x2 which results in
      (lt(x2,x3)→^*true⟹x3→^*x5⟹IF(true,x2,x5) > 1024_1^1(s(x2),x5))
    • Applying Rule "Variable in Equation" allows to substitute x5 by x3 which results in
      (lt(x2,x3)→^*true⟹IF(true,x2,x3) > 1024_1^1(s(x2),x3))
    • Applying Rule "Induction" on lt(x2,x3)→^*true results in the following 3 new constraints.
      1. (true→^*true⟹IF(true,0,s(x12)) > 1024_1^1(s(0),s(x12)))
         • Applying Rule "Same Constructor" results in
           IF(true,0,s(x12)) > 1024_1^1(s(0),s(x12))

         • This constraint is kept as final constraint.

      2. (lt(x14,x13)→^*true⟹(lt(x14,x13)→^*true⟹IF(true,x14,x13) > 1024_1^1(s(x14),x13))⟹IF(true,s(x14),s(x13)) > 1024_1^1(s(s(x14)),s(x13)))
         • Applying Rule "Simplify Conditions" results in
           (IF(true,x14,x13) > 1024_1^1(s(x14),x13)⟹IF(true,s(x14),s(x13)) > 1024_1^1(s(s(x14)),s(x13)))

         • This constraint is kept as final constraint.

      3. (false→^*true⟹IF(true,x15,0) > 1024_1^1(s(x15),0))
         • Applying Rule "Different Constructors" allows to drop this constraint.
  • For the chain we build the initial constraint

    (1024_1^1(s(x6),x7)→^*1024_1^1(x8,x9)⟹1024_1^1(x8,x9)≥IF(lt(x8,x9),x8,x9))

    which is simplified as follows.

    • Applying Rule "Same Constructor" results in
      (s(x6)→^*x8⟹x7→^*x9⟹1024_1^1(x8,x9)≥IF(lt(x8,x9),x8,x9))
    • Applying Rule "Variable in Equation" allows to substitute x8 by s(x6) which results in
      (x7→^*x9⟹1024_1^1(s(x6),x9)≥IF(lt(s(x6),x9),s(x6),x9))
    • Applying Rule "Variable in Equation" allows to substitute x9 by x7 which results in
      1024_1^1(s(x6),x7)≥IF(lt(s(x6),x7),s(x6),x7)

    • This constraint is kept as final constraint.

  We remove those pairs which are strictly decreasing and bounded.

  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.

  1024_1^1(x,x_removed)	→	IF(lt(x,x_removed),x,x_removed)	(32)
  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

  lt#(s(x),s(y))

  →

  lt#(x,y)

  (17)

  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.

  lt#(s(x),s(y))	→	lt#(x,y)	(17)
  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

  double#(s(x))

  →

  double#(x)

  (18)

  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.

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

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