Termination Proof using AProVE

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)

Termination of R to be shown.

     ↳Removing Redundant Rules

Removing the following rules from R which fullfill a polynomial ordering:

old(free(x)) -> free(old(x))

where the Polynomial interpretation:

POL(serve) = 1

POL(top1(x₁, x₂)) = x₁ + x₂

POL(old(x₁)) = 2·x₁

POL(free(x₁)) = 1 + x₁

POL(check(x₁)) = x₁

POL(top2(x₁, x₂)) = x₁ + x₂

POL(new(x₁)) = 1 + x₁

was used.

Not all Rules of R can be deleted, so we still have to regard a part of R.

     ↳RRRPolo

       →TRS2

         ↳Removing Redundant Rules

Removing the following rules from R which fullfill a polynomial ordering:

old(serve) -> free(serve)

where the Polynomial interpretation:

POL(serve) = 1

POL(top1(x₁, x₂)) = x₁ + x₂

POL(old(x₁)) = 2·x₁

POL(free(x₁)) = x₁

POL(check(x₁)) = x₁

POL(top2(x₁, x₂)) = x₁ + x₂

POL(new(x₁)) = x₁

was used.

Not all Rules of R can be deleted, so we still have to regard a part of R.

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains three SCCs.

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 1

                 ↳Size-Change Principle

Dependency Pair:

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

Rules:

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

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

free(x₁) -> free(x₁)

We obtain no new DP problems.

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 2

                 ↳Size-Change Principle

Dependency Pairs:

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

Rules:

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

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{3, 2, 1}	,	{3, 2, 1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{3, 2, 1}	,	{3, 2, 1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

old(x₁) -> old(x₁)
free(x₁) -> free(x₁)
new(x₁) -> new(x₁)

We obtain no new DP problems.

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 3

                 ↳Modular Removal of Rules

Dependency Pairs:

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

Rules:

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

We have the following set of usable rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(serve) = 0

POL(TOP2(x₁, x₂)) = x₁ + x₂

POL(old(x₁)) = x₁

POL(free(x₁)) = x₁

POL(check(x₁)) = x₁

POL(TOP1(x₁, x₂)) = x₁ + x₂

POL(new(x₁)) = x₁

We have the following set D of usable symbols: {serve, TOP2, old, check, free, TOP1, new}
No Dependency Pairs can be deleted.
8 non usable rules have been deleted.

The result of this processor delivers one new DP problem.

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

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

The transformation is resulting in one new DP problem:

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

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

The transformation is resulting in one new DP problem:

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 6

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

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

The transformation is resulting in one new DP problem:

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

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

The transformation is resulting in one new DP problem:

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

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

The transformation is resulting in one new DP problem:

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 9

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

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

The transformation is resulting in one new DP problem:

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 10

                 ↳Negative Polynomial Order

Dependency Pairs:

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

The following Dependency Pairs can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

check(free(x)) -> free(check(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
new(serve) -> free(serve)
new(free(x)) -> free(new(x))

Used ordering:
Polynomial Order with Interpretation:

POL( TOP1(x₁, x₂) ) = x₁ + x₂

POL( free(x₁) ) = x₁

POL( serve ) = 1

POL( TOP2(x₁, x₂) ) = x₁ + x₂

POL( check(x₁) ) = 0

POL( new(x₁) ) = x₁

POL( old(x₁) ) = 0

This results in one new DP problem.

     ↳RRRPolo

       →TRS2

         ↳RRRPolo

           →TRS3

             ↳DPs

...

               →DP Problem 11

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

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

Rules:

new(serve) -> free(serve)
new(free(x)) -> free(new(x))
check(old(x)) -> old(x)
check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(free(x)) -> free(check(x))

The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes

POL(serve)	= 1
POL(top1(x₁, x₂))	= x₁ + x₂
POL(old(x₁))	= 2·x₁
POL(free(x₁))	= 1 + x₁
POL(check(x₁))	= x₁
POL(top2(x₁, x₂))	= x₁ + x₂
POL(new(x₁))	= 1 + x₁