Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 MRRProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 UsableRulesProof (⇔)
    ↳14 QDP
    ↳15 MNOCProof (⇔)
    ↳16 QDP
    ↳17 MRRProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 Narrowing (⇔)
    ↳25 QDP
    ↳26 DependencyGraphProof (⇔)
    ↳27 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
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)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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)
MAT(f(x), f(y)) → F(mat(x, y))
MAT(f(x), f(y)) → MAT(x, y)
CHK(no(c)) → ACTIVE(c)
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)

The TRS R consists of the following 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)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 27 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)

The TRS R consists of the following 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)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)

The TRS R consists of the following rules:

f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F(active(x)) → ACTIVE(f(x))
ACTIVE(f(x)) → F(f(x))
F(active(x)) → F(x)
F(no(x)) → F(x)
F(mark(x)) → F(x)

Strictly oriented rules of the TRS R:

active(f(x)) → mark(f(f(x)))

Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = 1 + x1   
POL(F(x1)) = x1   
POL(active(x1)) = 2 + x1   
POL(f(x1)) = x1   
POL(mark(x1)) = 1 + x1   
POL(no(x1)) = 1 + x1   

(9) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following 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)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)

The set Q consists of the following terms:

mat(f(x0), f(y))
mat(f(x0), c)

We have to consider all minimal (P,Q,R)-chains.

(17) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

CHK(no(f(x))) → CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

Strictly oriented rules of the TRS R:

mat(f(x), c) → no(c)

Used ordering: Polynomial interpretation [POLO]:

POL(CHK(x1)) = x1   
POL(X) = 0   
POL(c) = 2   
POL(f(x1)) = 2·x1   
POL(mat(x1, x2)) = x1 + 2·x2   
POL(no(x1)) = 1 + x1   
POL(y) = 0   

(18) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))

The set Q consists of the following terms:

mat(f(x0), f(y))
mat(f(x0), c)

We have to consider all minimal (P,Q,R)-chains.

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) TRUE

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following 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)))
mat(f(x), f(y)) → f(mat(x, y))
chk(no(c)) → active(c)
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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) → active(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(24) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule TP(mark(x)) → TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))) at position [0] we obtained the following new rules [LPAR04]:

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)))

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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 TRS R consists of the following rules:

mat(f(x), f(y)) → f(mat(x, y))
mat(f(x), c) → no(c)
chk(no(f(x))) → f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) → active(c)
f(active(x)) → active(f(x))
f(no(x)) → no(f(x))
f(mark(x)) → mark(f(x))
active(f(x)) → mark(f(f(x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(27) TRUE