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

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 QTRSRRRProof (⇔)
    ↳6 QTRS
    ↳7 DependencyPairsProof (⇔)
    ↳8 QDP
    ↳9 DependencyGraphProof (⇔)
    ↳10 AND
      ↳11 QDP
        ↳12 UsableRulesProof (⇔)
          ↳13 QDP
          ↳14 UsableRulesReductionPairsProof (⇔)
          ↳15 QDP
          ↳16 PisEmptyProof (⇔)
          ↳17 TRUE
        ↳18 UsableRulesProof (⇔)
          ↳19 QDP
      ↳20 QDP
        ↳21 UsableRulesProof (⇔)
          ↳22 QDP
          ↳23 UsableRulesReductionPairsProof (⇔)
          ↳24 QDP
          ↳25 PisEmptyProof (⇔)
          ↳26 TRUE
        ↳27 UsableRulesProof (⇔)
          ↳28 QDP
      ↳29 QDP
        ↳30 UsableRulesProof (⇔)
          ↳31 QDP
          ↳32 UsableRulesReductionPairsProof (⇔)
          ↳33 QDP
          ↳34 PisEmptyProof (⇔)
          ↳35 TRUE
        ↳36 UsableRulesProof (⇔)
          ↳37 QDP
      ↳38 QDP
        ↳39 UsableRulesProof (⇔)
          ↳40 QDP
        ↳41 UsableRulesProof (⇔)
          ↳42 QDP
          ↳43 UsableRulesReductionPairsProof (⇔)
          ↳44 QDP
          ↳45 PisEmptyProof (⇔)
          ↳46 TRUE
      ↳47 QDP
        ↳48 UsableRulesProof (⇔)
          ↳49 QDP
        ↳50 UsableRulesProof (⇔)
          ↳51 QDP
          ↳52 UsableRulesReductionPairsProof (⇔)
          ↳53 QDP
          ↳54 PisEmptyProof (⇔)
          ↳55 TRUE
      ↳56 QDP
        ↳57 MRRProof (⇔)
        ↳58 QDP
        ↳59 QDPOrderProof (⇔)
        ↳60 QDP
        ↳61 UsableRulesProof (⇔)
        ↳62 QDP
        ↳63 UsableRulesReductionPairsProof (⇔)
        ↳64 QDP
        ↳65 DependencyGraphProof (⇔)
        ↳66 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(active(x))) → f(g(f(c(mark(x)))))
c(active(x)) → d(mark(x))
h(active(x)) → d(c(mark(x)))
f(mark(x)) → mark(f(active(x)))
c(mark(x)) → c(active(x))
g(mark(x)) → g(active(x))
d(mark(x)) → d(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(c(x)) → c(x)
active(c(x)) → c(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(d(x)) → d(x)
active(d(x)) → d(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(active(x))) → f(g(f(c(mark(x)))))
c(active(x)) → d(mark(x))
h(active(x)) → d(c(mark(x)))
f(mark(x)) → mark(f(active(x)))
c(mark(x)) → c(active(x))
g(mark(x)) → g(active(x))
d(mark(x)) → d(active(x))
h(mark(x)) → mark(h(active(x)))
mark(f(x)) → f(x)
active(f(x)) → f(x)
mark(c(x)) → c(x)
active(c(x)) → c(x)
mark(g(x)) → g(x)
active(g(x)) → g(x)
mark(d(x)) → d(x)
active(d(x)) → d(x)
mark(h(x)) → h(x)
active(h(x)) → h(x)

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(active(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = 1 + x1   
POL(mark(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

active(h(X)) → mark(c(d(X)))


(6) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
ACTIVE(f(f(X))) → C(f(g(f(X))))
ACTIVE(f(f(X))) → F(g(f(X)))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(c(X)) → MARK(d(X))
ACTIVE(c(X)) → D(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(X)) → MARK(X)
MARK(c(X)) → ACTIVE(c(X))
MARK(g(X)) → ACTIVE(g(X))
MARK(d(X)) → ACTIVE(d(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → H(mark(X))
MARK(h(X)) → MARK(X)
F(mark(X)) → F(X)
F(active(X)) → F(X)
C(mark(X)) → C(X)
C(active(X)) → C(X)
G(mark(X)) → G(X)
G(active(X)) → G(X)
D(mark(X)) → D(X)
D(active(X)) → D(X)
H(mark(X)) → H(X)
H(active(X)) → H(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 6 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(12) 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.

(13) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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

(14) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

H(active(X)) → H(X)
H(mark(X)) → H(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(H(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(15) Obligation:

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) 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.

(19) Obligation:

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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

(20) Obligation:

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

D(active(X)) → D(X)
D(mark(X)) → D(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(21) 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.

(22) Obligation:

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

D(active(X)) → D(X)
D(mark(X)) → D(X)

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

(23) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

D(active(X)) → D(X)
D(mark(X)) → D(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(D(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(24) Obligation:

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) 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.

(28) Obligation:

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

D(active(X)) → D(X)
D(mark(X)) → D(X)

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

(29) Obligation:

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(30) 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.

(31) Obligation:

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

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

(32) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

G(active(X)) → G(X)
G(mark(X)) → G(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(G(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(33) Obligation:

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) 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.

(37) Obligation:

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

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

(38) Obligation:

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

C(active(X)) → C(X)
C(mark(X)) → C(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(39) 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.

(40) Obligation:

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

C(active(X)) → C(X)
C(mark(X)) → C(X)

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

(41) 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.

(42) Obligation:

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

C(active(X)) → C(X)
C(mark(X)) → C(X)

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

(43) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

C(active(X)) → C(X)
C(mark(X)) → C(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(C(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(44) Obligation:

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(48) 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.

(49) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

(50) 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.

(51) Obligation:

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

F(active(X)) → F(X)
F(mark(X)) → F(X)

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

(52) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F(active(X)) → F(X)
F(mark(X)) → F(X)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F(x1)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(mark(x1)) = 2·x1   

(53) Obligation:

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

(54) PisEmptyProof (EQUIVALENT transformation)

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

(55) TRUE

(56) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(f(X)) → MARK(X)
MARK(c(X)) → ACTIVE(c(X))
ACTIVE(c(X)) → MARK(d(X))
MARK(g(X)) → ACTIVE(g(X))
MARK(d(X)) → ACTIVE(d(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(h(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(57) 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:

MARK(f(X)) → MARK(X)
MARK(h(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(f(x1)) = 1 + x1   
POL(g(x1)) = x1   
POL(h(x1)) = 1 + x1   
POL(mark(x1)) = x1   

(58) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
ACTIVE(c(X)) → MARK(d(X))
MARK(g(X)) → ACTIVE(g(X))
MARK(d(X)) → ACTIVE(d(X))
MARK(h(X)) → ACTIVE(h(mark(X)))

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(g(X)) → ACTIVE(g(X))
MARK(h(X)) → ACTIVE(h(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ACTIVE(x1) ) = 1


POL( f(x1) ) = x1 + 1


POL( h(x1) ) = x1 + 1


POL( mark(x1) ) = 1


POL( g(x1) ) = 1


POL( active(x1) ) = 1


POL( c(x1) ) = 0


POL( d(x1) ) = 0


POL( MARK(x1) ) = x1 + 1



The following usable rules [FROCOS05] were oriented:

d(active(X)) → d(X)
d(mark(X)) → d(X)
c(active(X)) → c(X)
c(mark(X)) → c(X)

(60) Obligation:

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

ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
ACTIVE(c(X)) → MARK(d(X))
MARK(d(X)) → ACTIVE(d(X))

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
mark(f(X)) → active(f(mark(X)))
mark(c(X)) → active(c(X))
mark(g(X)) → active(g(X))
mark(d(X)) → active(d(X))
mark(h(X)) → active(h(mark(X)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
c(mark(X)) → c(X)
c(active(X)) → c(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
d(mark(X)) → d(X)
d(active(X)) → d(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

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

(61) 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.

(62) Obligation:

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

ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
ACTIVE(c(X)) → MARK(d(X))
MARK(d(X)) → ACTIVE(d(X))

The TRS R consists of the following rules:

d(active(X)) → d(X)
d(mark(X)) → d(X)
c(active(X)) → c(X)
c(mark(X)) → c(X)
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(active(X)) → g(X)
g(mark(X)) → g(X)

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

(63) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

d(active(X)) → d(X)
d(mark(X)) → d(X)
c(active(X)) → c(X)
c(mark(X)) → c(X)
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(mark(X)) → g(X)
g(active(X)) → g(X)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(mark(x1)) = 2·x1   

(64) Obligation:

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

ACTIVE(f(f(X))) → MARK(c(f(g(f(X)))))
MARK(c(X)) → ACTIVE(c(X))
ACTIVE(c(X)) → MARK(d(X))
MARK(d(X)) → ACTIVE(d(X))

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

(65) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.

(66) TRUE