Term Rewriting System R:
[x, y]
top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TOP1(free(x), y) -> TOP2(check(new(x)), y)
TOP1(free(x), y) -> CHECK(new(x))
TOP1(free(x), y) -> NEW(x)
TOP1(free(x), y) -> TOP2(new(x), check(y))
TOP1(free(x), y) -> CHECK(y)
TOP1(free(x), y) -> TOP2(check(x), new(y))
TOP1(free(x), y) -> CHECK(x)
TOP1(free(x), y) -> NEW(y)
TOP1(free(x), y) -> TOP2(x, check(new(y)))
TOP1(free(x), y) -> CHECK(new(y))
TOP2(x, free(y)) -> TOP1(check(new(x)), y)
TOP2(x, free(y)) -> CHECK(new(x))
TOP2(x, free(y)) -> NEW(x)
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP2(x, free(y)) -> CHECK(y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP2(x, free(y)) -> CHECK(x)
TOP2(x, free(y)) -> NEW(y)
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP2(x, free(y)) -> CHECK(new(y))
NEW(free(x)) -> NEW(x)
OLD(free(x)) -> OLD(x)
CHECK(free(x)) -> CHECK(x)
CHECK(new(x)) -> NEW(check(x))
CHECK(new(x)) -> CHECK(x)
CHECK(old(x)) -> OLD(check(x))
CHECK(old(x)) -> CHECK(x)

Furthermore, R contains five SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar


Dependency Pair:

NEW(free(x)) -> NEW(x)


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




The following dependency pair can be strictly oriented:

NEW(free(x)) -> NEW(x)


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
NEW(x1) -> NEW(x1)
free(x1) -> free(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar


Dependency Pair:

OLD(free(x)) -> OLD(x)


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




The following dependency pair can be strictly oriented:

OLD(free(x)) -> OLD(x)


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
OLD(x1) -> OLD(x1)
free(x1) -> free(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar


Dependency Pair:

CHECK(free(x)) -> CHECK(x)


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




The following dependency pair can be strictly oriented:

CHECK(free(x)) -> CHECK(x)


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
CHECK(x1) -> CHECK(x1)
free(x1) -> free(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 8
             ↳Dependency Graph
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳Argument Filtering and Ordering
       →DP Problem 5
         ↳Nar


Dependency Pairs:

CHECK(old(x)) -> CHECK(x)
CHECK(new(x)) -> CHECK(x)


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




The following dependency pairs can be strictly oriented:

CHECK(old(x)) -> CHECK(x)
CHECK(new(x)) -> CHECK(x)


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
CHECK(x1) -> CHECK(x1)
new(x1) -> new(x1)
old(x1) -> old(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
           →DP Problem 9
             ↳Dependency Graph
       →DP Problem 5
         ↳Nar


Dependency Pair:


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Narrowing Transformation


Dependency Pairs:

TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(x), y) -> TOP2(x, check(new(y)))
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(x), y) -> TOP2(check(x), new(y))
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP1(free(x), y) -> TOP2(new(x), check(y))
TOP2(x, free(y)) -> TOP1(check(new(x)), y)
TOP1(free(x), y) -> TOP2(check(new(x)), y)


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP1(free(x), y) -> TOP2(check(new(x)), y)
three new Dependency Pairs are created:

TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(x), y) -> TOP2(x, check(new(y)))
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP1(free(x), y) -> TOP2(check(x), new(y))
TOP2(x, free(y)) -> TOP1(check(new(x)), y)
TOP1(free(x), y) -> TOP2(new(x), check(y))
TOP2(x, free(y)) -> TOP1(x, check(new(y)))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP1(free(x), y) -> TOP2(new(x), check(y))
six new Dependency Pairs are created:

TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 11
                 ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(x), y) -> TOP2(x, check(new(y)))
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP1(free(x), y) -> TOP2(check(x), new(y))
TOP2(x, free(y)) -> TOP1(check(new(x)), y)
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP1(free(x), y) -> TOP2(check(x), new(y))
six new Dependency Pairs are created:

TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP1(free(x), serve) -> TOP2(check(x), free(serve))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 12
                 ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(x), serve) -> TOP2(check(x), free(serve))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP1(free(x), y) -> TOP2(x, check(new(y)))
TOP2(x, free(y)) -> TOP1(check(new(x)), y)
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP1(free(x), y) -> TOP2(x, check(new(y)))
three new Dependency Pairs are created:

TOP1(free(x), y') -> TOP2(x, new(check(y')))
TOP1(free(x), free(x'')) -> TOP2(x, check(free(new(x''))))
TOP1(free(x), serve) -> TOP2(x, check(free(serve)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 13
                 ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(x), serve) -> TOP2(x, check(free(serve)))
TOP1(free(x), free(x'')) -> TOP2(x, check(free(new(x''))))
TOP1(free(x), y') -> TOP2(x, new(check(y')))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(check(new(x)), y)
TOP1(free(x), serve) -> TOP2(check(x), free(serve))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP2(x, free(y)) -> TOP1(check(new(x)), y)
three new Dependency Pairs are created:

TOP2(x'', free(y)) -> TOP1(new(check(x'')), y)
TOP2(free(x''), free(y)) -> TOP1(check(free(new(x''))), y)
TOP2(serve, free(y)) -> TOP1(check(free(serve)), y)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 14
                 ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(x), free(x'')) -> TOP2(x, check(free(new(x''))))
TOP1(free(x), y') -> TOP2(x, new(check(y')))
TOP1(free(x), serve) -> TOP2(check(x), free(serve))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP2(serve, free(y)) -> TOP1(check(free(serve)), y)
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP2(free(x''), free(y)) -> TOP1(check(free(new(x''))), y)
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP2(x'', free(y)) -> TOP1(new(check(x'')), y)
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(new(x), check(y))
TOP1(free(x), serve) -> TOP2(x, check(free(serve)))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP2(x, free(y)) -> TOP1(new(x), check(y))
six new Dependency Pairs are created:

TOP2(free(x''), free(y)) -> TOP1(free(new(x'')), check(y))
TOP2(serve, free(y)) -> TOP1(free(serve), check(y))
TOP2(x, free(free(x''))) -> TOP1(new(x), free(check(x'')))
TOP2(x, free(new(x''))) -> TOP1(new(x), new(check(x'')))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(check(x'')))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(x''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 15
                 ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(x), serve) -> TOP2(x, check(free(serve)))
TOP1(free(x), serve) -> TOP2(check(x), free(serve))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP1(free(x), y') -> TOP2(x, new(check(y')))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(x''))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(check(x'')))
TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP2(x, free(new(x''))) -> TOP1(new(x), new(check(x'')))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP2(serve, free(y)) -> TOP1(free(serve), check(y))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP2(serve, free(y)) -> TOP1(check(free(serve)), y)
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP2(x, free(free(x''))) -> TOP1(new(x), free(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP2(free(x''), free(y)) -> TOP1(free(new(x'')), check(y))
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP2(free(x''), free(y)) -> TOP1(check(free(new(x''))), y)
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP2(x'', free(y)) -> TOP1(new(check(x'')), y)
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(check(x), new(y))
TOP1(free(x), free(x'')) -> TOP2(x, check(free(new(x''))))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP2(x, free(y)) -> TOP1(check(x), new(y))
six new Dependency Pairs are created:

TOP2(free(x''), free(y)) -> TOP1(free(check(x'')), new(y))
TOP2(new(x''), free(y)) -> TOP1(new(check(x'')), new(y))
TOP2(old(x''), free(y)) -> TOP1(old(check(x'')), new(y))
TOP2(old(x''), free(y)) -> TOP1(old(x''), new(y))
TOP2(x, free(free(x''))) -> TOP1(check(x), free(new(x'')))
TOP2(x, free(serve)) -> TOP1(check(x), free(serve))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 16
                 ↳Narrowing Transformation


Dependency Pairs:

TOP1(free(x), serve) -> TOP2(check(x), free(serve))
TOP1(free(x), free(x'')) -> TOP2(x, check(free(new(x''))))
TOP2(x, free(serve)) -> TOP1(check(x), free(serve))
TOP1(free(x), y') -> TOP2(x, new(check(y')))
TOP2(x, free(free(x''))) -> TOP1(check(x), free(new(x'')))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP2(old(x''), free(y)) -> TOP1(old(check(x'')), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP2(new(x''), free(y)) -> TOP1(new(check(x'')), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP2(free(x''), free(y)) -> TOP1(free(check(x'')), new(y))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(x''))
TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(check(x'')))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP2(x, free(new(x''))) -> TOP1(new(x), new(check(x'')))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP2(serve, free(y)) -> TOP1(free(serve), check(y))
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP2(x, free(free(x''))) -> TOP1(new(x), free(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP2(free(x''), free(y)) -> TOP1(free(new(x'')), check(y))
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP2(serve, free(y)) -> TOP1(check(free(serve)), y)
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP2(free(x''), free(y)) -> TOP1(check(free(new(x''))), y)
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(x'', free(y)) -> TOP1(new(check(x'')), y)
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x, free(y)) -> TOP1(x, check(new(y)))
TOP1(free(x), serve) -> TOP2(x, check(free(serve)))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TOP2(x, free(y)) -> TOP1(x, check(new(y)))
three new Dependency Pairs are created:

TOP2(x, free(y')) -> TOP1(x, new(check(y')))
TOP2(x, free(free(x''))) -> TOP1(x, check(free(new(x''))))
TOP2(x, free(serve)) -> TOP1(x, check(free(serve)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
       →DP Problem 4
         ↳AFS
       →DP Problem 5
         ↳Nar
           →DP Problem 10
             ↳Nar
             ...
               →DP Problem 17
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

TOP2(x, free(serve)) -> TOP1(x, check(free(serve)))
TOP1(free(x), serve) -> TOP2(x, check(free(serve)))
TOP2(x, free(free(x''))) -> TOP1(x, check(free(new(x''))))
TOP1(free(x), free(x'')) -> TOP2(x, check(free(new(x''))))
TOP2(x, free(y')) -> TOP1(x, new(check(y')))
TOP1(free(x), y') -> TOP2(x, new(check(y')))
TOP2(x, free(serve)) -> TOP1(check(x), free(serve))
TOP1(free(x), free(x'')) -> TOP2(check(x), free(new(x'')))
TOP2(x, free(free(x''))) -> TOP1(check(x), free(new(x'')))
TOP1(free(old(x'')), y) -> TOP2(old(x''), new(y))
TOP2(old(x''), free(y)) -> TOP1(old(check(x'')), new(y))
TOP1(free(old(x'')), y) -> TOP2(old(check(x'')), new(y))
TOP2(new(x''), free(y)) -> TOP1(new(check(x'')), new(y))
TOP1(free(new(x'')), y) -> TOP2(new(check(x'')), new(y))
TOP2(free(x''), free(y)) -> TOP1(free(check(x'')), new(y))
TOP1(free(free(x'')), y) -> TOP2(free(check(x'')), new(y))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(x''))
TOP1(free(x), old(x'')) -> TOP2(new(x), old(check(x'')))
TOP2(x, free(old(x''))) -> TOP1(new(x), old(check(x'')))
TOP1(free(x), new(x'')) -> TOP2(new(x), new(check(x'')))
TOP2(serve, free(y)) -> TOP1(free(serve), check(y))
TOP1(free(x), free(x'')) -> TOP2(new(x), free(check(x'')))
TOP2(x, free(new(x''))) -> TOP1(new(x), new(check(x'')))
TOP1(free(serve), y) -> TOP2(free(serve), check(y))
TOP2(x, free(free(x''))) -> TOP1(new(x), free(check(x'')))
TOP1(free(free(x'')), y) -> TOP2(free(new(x'')), check(y))
TOP2(free(x''), free(y)) -> TOP1(free(new(x'')), check(y))
TOP1(free(serve), y) -> TOP2(check(free(serve)), y)
TOP2(serve, free(y)) -> TOP1(check(free(serve)), y)
TOP1(free(free(x'')), y) -> TOP2(check(free(new(x''))), y)
TOP2(free(x''), free(y)) -> TOP1(check(free(new(x''))), y)
TOP1(free(x''), y) -> TOP2(new(check(x'')), y)
TOP2(x'', free(y)) -> TOP1(new(check(x'')), y)
TOP1(free(x), serve) -> TOP2(check(x), free(serve))


Rules:


top1(free(x), y) -> top2(check(new(x)), y)
top1(free(x), y) -> top2(new(x), check(y))
top1(free(x), y) -> top2(check(x), new(y))
top1(free(x), y) -> top2(x, check(new(y)))
top2(x, free(y)) -> top1(check(new(x)), y)
top2(x, free(y)) -> top1(new(x), check(y))
top2(x, free(y)) -> top1(check(x), new(y))
top2(x, free(y)) -> top1(x, check(new(y)))
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
old(free(x)) -> free(old(x))
old(serve) -> free(serve)
check(free(x)) -> free(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(check(x))
check(old(x)) -> old(x)


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:41 minutes