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)))

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

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

F(no(x)) -> F(x)

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(chk(x₁)) = 0

POL(active(x₁)) = x₁

POL(ACTIVE(x₁)) = x₁

POL(c) = 0

POL(X) = 0

POL(tp(x₁)) = 0

POL(no(x₁)) = 1 + x₁

POL(mark(x₁)) = x₁

POL(y) = 1

POL(f(x₁)) = x₁

POL(F(x₁)) = x₁

POL(mat(x₁, x₂)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

F(mark(x)) -> F(x)
F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))
ACTIVE(f(x)) -> F(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)))

The following dependency pairs can be strictly oriented:

F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(chk(x₁)) = x₁

POL(active(x₁)) = 1 + x₁

POL(ACTIVE(x₁)) = x₁

POL(c) = 0

POL(X) = 0

POL(tp(x₁)) = 0

POL(no(x₁)) = 1

POL(mark(x₁)) = x₁

POL(y) = 1

POL(f(x₁)) = x₁

POL(F(x₁)) = x₁

POL(mat(x₁, x₂)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining

Dependency Pairs:

F(mark(x)) -> F(x)
ACTIVE(f(x)) -> F(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)))

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

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 6

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining

Dependency Pair:

F(mark(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)))

The following dependency pair can be strictly oriented:

F(mark(x)) -> F(x)

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(chk(x₁)) = x₁

POL(active(x₁)) = 1 + x₁

POL(c) = 0

POL(X) = 0

POL(tp(x₁)) = 0

POL(no(x₁)) = 1

POL(mark(x₁)) = 1 + x₁

POL(y) = 1

POL(f(x₁)) = x₁

POL(F(x₁)) = 1 + x₁

POL(mat(x₁, x₂)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Polo

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Remaining

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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(chk(x₁)) = x₁

POL(active(x₁)) = 0

POL(c) = 0

POL(X) = 0

POL(tp(x₁)) = 0

POL(no(x₁)) = x₁

POL(mark(x₁)) = 0

POL(y) = 1

POL(f(x₁)) = 1 + x₁

POL(CHK(x₁)) = 1 + x₁

POL(mat(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

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)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:
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)))

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

POL(chk(x₁))	= 0
POL(active(x₁))	= x₁
POL(ACTIVE(x₁))	= x₁
POL(c)	= 0
POL(X)	= 0
POL(tp(x₁))	= 0
POL(no(x₁))	= 1 + x₁
POL(mark(x₁))	= x₁
POL(y)	= 1
POL(f(x₁))	= x₁
POL(F(x₁))	= x₁
POL(mat(x₁, x₂))	= 1