Termination Proof using AProVE

Term Rewriting System R:
[x]
top(free(x)) -> top(check(new(x)))
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.

TRS

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

TOP(free(x)) -> TOP(check(new(x)))
TOP(free(x)) -> CHECK(new(x))
TOP(free(x)) -> NEW(x)
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.

TRS

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

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

Rules:

top(free(x)) -> top(check(new(x)))
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

As we are in the innermost case, we can delete all 9 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 6

             ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(NEW(x₁)) = x₁

POL(free(x₁)) = x₁

We have the following set D of usable symbols: {NEW}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

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

Rules:

top(free(x)) -> top(check(new(x)))
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

As we are in the innermost case, we can delete all 9 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 7

             ↳Modular Removal of Rules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(free(x₁)) = x₁

POL(OLD(x₁)) = x₁

We have the following set D of usable symbols: {OLD}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

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

Rules:

top(free(x)) -> top(check(new(x)))
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

As we are in the innermost case, we can delete all 9 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳Modular Removal of Rules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(free(x₁)) = x₁

POL(CHECK(x₁)) = x₁

We have the following set D of usable symbols: {CHECK}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳Usable Rules (Innermost)

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

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

Rules:

top(free(x)) -> top(check(new(x)))
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

As we are in the innermost case, we can delete all 9 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 9

             ↳Modular Removal of Rules

       →DP Problem 5

         ↳UsableRules

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(old(x₁)) = x₁

POL(CHECK(x₁)) = x₁

POL(new(x₁)) = x₁

We have the following set D of usable symbols: {CHECK}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳Usable Rules (Innermost)

Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))

Rules:

top(free(x)) -> top(check(new(x)))
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

As we are in the innermost case, we can delete all 1 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳Modular Removal of Rules

Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

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

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(old(x₁)) = 1 + x₁

POL(free(x₁)) = x₁

POL(check(x₁)) = x₁

POL(TOP(x₁)) = x₁

POL(new(x₁)) = x₁

We have the following set D of usable symbols: {serve, old, free, check, TOP, new}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

old(serve) -> free(serve)

The result of this processor delivers one new DP problem.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 11

                 ↳Modular Removal of Rules

Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
old(free(x)) -> free(old(x))
new(serve) -> free(serve)
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(old(x₁)) = 2·x₁

POL(free(x₁)) = 1 + x₁

POL(check(x₁)) = x₁

POL(TOP(x₁)) = 1 + x₁

POL(new(x₁)) = 1 + x₁

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

The result of this processor delivers one new DP problem.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 12

                 ↳Modular Removal of Rules

Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

check(old(x)) -> old(check(x))
check(new(x)) -> new(check(x))
check(old(x)) -> old(x)
new(free(x)) -> free(new(x))
new(serve) -> free(serve)
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(old(x₁)) = 1 + x₁

POL(free(x₁)) = 2·x₁

POL(check(x₁)) = 2·x₁

POL(TOP(x₁)) = x₁

POL(new(x₁)) = x₁

check(old(x)) -> old(check(x))
check(old(x)) -> old(x)

The result of this processor delivers one new DP problem.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 13

                 ↳Narrowing Transformation

Dependency Pair:

TOP(free(x)) -> TOP(check(new(x)))

Rules:

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

Strategy:

innermost

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

TOP(free(x)) -> TOP(check(new(x)))

three new Dependency Pairs are created:

TOP(free(x'')) -> TOP(new(check(x'')))
TOP(free(free(x''))) -> TOP(check(free(new(x''))))
TOP(free(serve)) -> TOP(check(free(serve)))

The transformation is resulting in one new DP problem:

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 14

                 ↳Negative Polynomial Order

Dependency Pairs:

TOP(free(serve)) -> TOP(check(free(serve)))
TOP(free(free(x''))) -> TOP(check(free(new(x''))))
TOP(free(x'')) -> TOP(new(check(x'')))

Rules:

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

Strategy:

innermost

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

TOP(free(serve)) -> TOP(check(free(serve)))

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x₁) ) = x₁

POL( free(x₁) ) = x₁

POL( serve ) = 1

POL( check(x₁) ) = 0

POL( new(x₁) ) = x₁

This results in one new DP problem.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 15

                 ↳Semantic Labelling

Dependency Pairs:

TOP(free(free(x''))) -> TOP(check(free(new(x''))))
TOP(free(x'')) -> TOP(new(check(x'')))

Rules:

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

Strategy:

innermost

Using Semantic Labelling, the DP problem could be transformed. The following model was found:

TOP(x₀) = 1

free(x₀) = x₀

new(x₀) = x₀

check(x₀) = 1

serve = 0

From the dependency graph we obtain 1 (labeled) SCCs which each result in correspondingDP problem.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 16

                 ↳Modular Removal of Rules

Dependency Pairs:

TOP₁(free₁(x'')) -> TOP₁(new₁(check₁(x'')))
TOP₁(free₁(free₁(x''))) -> TOP₁(check₁(free₁(new₁(x''))))

Rules:

check₀(new₀(x)) -> new₁(check₀(x))
check₀(free₀(x)) -> free₁(check₀(x))
new₀(free₀(x)) -> free₀(new₀(x))
new₀(serve) -> free₀(serve)
new₁(free₁(x)) -> free₁(new₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(free₁(x)) -> free₁(check₁(x))

Strategy:

innermost

We have the following set of usable rules:

new₁(free₁(x)) -> free₁(new₁(x))
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(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(free_1(x₁)) = x₁

POL(check_1(x₁)) = x₁

POL(TOP_1(x₁)) = 1 + x₁

POL(new_1(x₁)) = x₁

We have the following set D of usable symbols: {free₁, check₁, TOP₁, new₁}
No Dependency Pairs can be deleted.
4 non usable rules have been deleted.

The result of this processor delivers one new DP problem.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

       →DP Problem 5

         ↳UsableRules

           →DP Problem 10

             ↳MRR

...

               →DP Problem 17

                 ↳Modular Removal of Rules

Dependency Pairs:

TOP₁(free₁(free₁(x''))) -> TOP₁(check₁(free₁(new₁(x''))))
TOP₁(free₁(x'')) -> TOP₁(new₁(check₁(x'')))

Rules:

new₁(free₁(x)) -> free₁(new₁(x))
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))

Strategy:

innermost

We have the following set of usable rules:

new₁(free₁(x)) -> free₁(new₁(x))
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(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(free_1(x₁)) = 1 + x₁

POL(check_1(x₁)) = x₁

POL(TOP_1(x₁)) = 1 + x₁

POL(new_1(x₁)) = x₁

We have the following set D of usable symbols: {free₁, check₁, TOP₁, new₁}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

TOP₁(free₁(free₁(x''))) -> TOP₁(check₁(free₁(new₁(x''))))
TOP₁(free₁(x'')) -> TOP₁(new₁(check₁(x'')))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

Innermost Termination of R successfully shown.
Duration:
0:15 minutes

POL(old(x₁))	= x₁
POL(CHECK(x₁))	= x₁
POL(new(x₁))	= x₁

POL(serve)	= 0
POL(old(x₁))	= 1 + x₁
POL(free(x₁))	= x₁
POL(check(x₁))	= x₁
POL(TOP(x₁))	= x₁
POL(new(x₁))	= x₁

POL(serve)	= 0
POL(old(x₁))	= 2·x₁
POL(free(x₁))	= 1 + x₁
POL(check(x₁))	= x₁
POL(TOP(x₁))	= 1 + x₁
POL(new(x₁))	= 1 + x₁

TOP(x₀)	= 1
free(x₀)	= x₀
new(x₀)	= x₀
check(x₀)	= 1
serve	= 0

POL(free_1(x₁))	= x₁
POL(check_1(x₁))	= x₁
POL(TOP_1(x₁))	= 1 + x₁
POL(new_1(x₁))	= x₁