Termination Proof using AProVE

Term Rewriting System R:
[X]
active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Innermost Termination of R to be shown.

TRS

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(c) -> F(g(c))
ACTIVE(c) -> G(c)
PROPER(f(X)) -> F(proper(X))
PROPER(f(X)) -> PROPER(X)
PROPER(g(X)) -> G(proper(X))
PROPER(g(X)) -> PROPER(X)
F(ok(X)) -> F(X)
G(ok(X)) -> G(X)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains four SCCs.

TRS

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

Dependency Pair:

F(ok(X)) -> F(X)

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(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 5

             ↳Modular Removal of Rules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

Dependency Pair:

F(ok(X)) -> F(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(ok(x₁)) = x₁

POL(F(x₁)) = x₁

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

F(ok(X)) -> F(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

Dependency Pair:

G(ok(X)) -> G(X)

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(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 6

             ↳Modular Removal of Rules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

Dependency Pair:

G(ok(X)) -> G(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(G(x₁)) = x₁

POL(ok(x₁)) = x₁

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

G(ok(X)) -> G(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

Dependency Pairs:

PROPER(g(X)) -> PROPER(X)
PROPER(f(X)) -> PROPER(X)

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(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 7

             ↳Modular Removal of Rules

       →DP Problem 4

         ↳UsableRules

Dependency Pairs:

PROPER(g(X)) -> PROPER(X)
PROPER(f(X)) -> PROPER(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(g(x₁)) = x₁

POL(PROPER(x₁)) = x₁

POL(f(x₁)) = x₁

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

PROPER(g(X)) -> PROPER(X)
PROPER(f(X)) -> PROPER(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)

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 8

             ↳Narrowing Transformation

Dependency Pairs:

TOP(ok(X)) -> TOP(active(X))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))

Strategy:

innermost

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

TOP(ok(X)) -> TOP(active(X))

two new Dependency Pairs are created:

TOP(ok(c)) -> TOP(mark(f(g(c))))
TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

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 8

             ↳Nar

...

               →DP Problem 9

                 ↳Narrowing Transformation

Dependency Pairs:

TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))
TOP(ok(c)) -> TOP(mark(f(g(c))))
TOP(mark(X)) -> TOP(proper(X))

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))

Strategy:

innermost

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

TOP(mark(X)) -> TOP(proper(X))

three new Dependency Pairs are created:

TOP(mark(c)) -> TOP(ok(c))
TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))

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 8

             ↳Nar

...

               →DP Problem 10

                 ↳Usable Rules (Innermost)

Dependency Pairs:

TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(ok(c)) -> TOP(mark(f(g(c))))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

Rules:

active(c) -> mark(f(g(c)))
active(f(g(X))) -> mark(g(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 8

             ↳Nar

...

               →DP Problem 11

                 ↳Negative Polynomial Order

Dependency Pairs:

TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(ok(c)) -> TOP(mark(f(g(c))))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

Rules:

proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))

Strategy:

innermost

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

TOP(ok(c)) -> TOP(mark(f(g(c))))

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

f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))

Used ordering:
Polynomial Order with Interpretation:

POL( TOP(x₁) ) = x₁

POL( ok(x₁) ) = x₁

POL( c ) = 1

POL( mark(x₁) ) = 0

POL( f(x₁) ) = 0

POL( g(x₁) ) = 0

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 8

             ↳Nar

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pairs:

TOP(mark(f(X''))) -> TOP(f(proper(X'')))
TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

Rules:

proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

TRS

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

       →DP Problem 4

         ↳UsableRules

           →DP Problem 8

             ↳Nar

...

               →DP Problem 13

                 ↳Modular Removal of Rules

Dependency Pairs:

TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

Rules:

proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))

Strategy:

innermost

We have the following set of usable rules:

f(ok(X)) -> ok(f(X))
proper(c) -> ok(c)
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
g(ok(X)) -> ok(g(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(proper(x₁)) = 2·x₁

POL(c) = 1

POL(g(x₁)) = x₁

POL(mark(x₁)) = 2·x₁

POL(TOP(x₁)) = x₁

POL(ok(x₁)) = x₁

POL(f(x₁)) = 2·x₁

We have the following set D of usable symbols: {proper, c, g, mark, TOP, ok, f}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

proper(c) -> ok(c)

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 8

             ↳Nar

...

               →DP Problem 14

                 ↳Modular Removal of Rules

Dependency Pairs:

TOP(mark(g(X''))) -> TOP(g(proper(X'')))
TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

Rules:

f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
g(ok(X)) -> ok(g(X))

Strategy:

innermost

We have the following set of usable rules:

f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
g(ok(X)) -> ok(g(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(proper(x₁)) = x₁

POL(g(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(TOP(x₁)) = x₁

POL(ok(x₁)) = x₁

POL(f(x₁)) = 1 + x₁

We have the following set D of usable symbols: {proper, g, mark, TOP, ok, f}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

TOP(ok(f(g(X'')))) -> TOP(mark(g(X'')))

No Rules can be 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 8

             ↳Nar

...

               →DP Problem 15

                 ↳Modular Removal of Rules

Dependency Pair:

TOP(mark(g(X''))) -> TOP(g(proper(X'')))

Rules:

f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
g(ok(X)) -> ok(g(X))

Strategy:

innermost

We have the following set of usable rules:

f(ok(X)) -> ok(f(X))
proper(f(X)) -> f(proper(X))
proper(g(X)) -> g(proper(X))
g(ok(X)) -> ok(g(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(proper(x₁)) = x₁

POL(g(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(TOP(x₁)) = 1 + x₁

POL(ok(x₁)) = x₁

POL(f(x₁)) = x₁

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

TOP(mark(g(X''))) -> TOP(g(proper(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:
0:12 minutes

POL(g(x₁))	= x₁
POL(PROPER(x₁))	= x₁
POL(f(x₁))	= x₁

POL(proper(x₁))	= 2·x₁
POL(c)	= 1
POL(g(x₁))	= x₁
POL(mark(x₁))	= 2·x₁
POL(TOP(x₁))	= x₁
POL(ok(x₁))	= x₁
POL(f(x₁))	= 2·x₁