Termination Proof using AProVE

Term Rewriting System R:
[x]
active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(f(x)) -> F(f(x))
CHK(no(f(x))) -> F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) -> F(f(f(f(f(f(X))))))
CHK(no(f(x))) -> F(f(f(f(f(X)))))
CHK(no(f(x))) -> F(f(f(f(X))))
CHK(no(f(x))) -> F(f(f(X)))
CHK(no(f(x))) -> F(f(X))
CHK(no(f(x))) -> F(X)
CHK(no(c)) -> ACTIVE(c)
MAT(f(x), f(y)) -> F(mat(x, y))
MAT(f(x), f(y)) -> MAT(x, y)
F(active(x)) -> ACTIVE(f(x))
F(active(x)) -> F(x)
F(no(x)) -> F(x)
F(mark(x)) -> F(x)
TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
TP(mark(x)) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(f(X))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(X)))))))
TP(mark(x)) -> F(f(f(f(f(f(X))))))
TP(mark(x)) -> F(f(f(f(f(X)))))
TP(mark(x)) -> F(f(f(f(X))))
TP(mark(x)) -> F(f(f(X)))
TP(mark(x)) -> F(f(X))
TP(mark(x)) -> F(X)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
F(active(x)) -> F(x)

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
F(active(x)) -> F(x)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

F(x₁) -> F(x₁)
active(x₁) -> active(x₁)
mark(x₁) -> mark(x₁)
no(x₁) -> no(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

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

CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

two new Dependency Pairs are created:

CHK(no(f(f(y)))) -> CHK(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))
CHK(no(f(c))) -> CHK(no(c))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 5

             ↳Narrowing Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

CHK(no(f(f(y)))) -> CHK(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

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

CHK(no(f(f(y)))) -> CHK(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))

no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pair:

TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

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

TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

two new Dependency Pairs are created:

TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
TP(mark(c)) -> TP(chk(no(c)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

TP(mark(c)) -> TP(chk(no(c)))
TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

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

TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

TP(mark(c)) -> TP(chk(no(c)))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes