YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z003-split.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 40 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 18 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 2 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 39 ms)
        ↳11 QDP
        ↳12 UsableRulesProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 PisEmptyProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QDPSizeChangeProof (⇔, 0 ms)
        ↳32 YES
    ↳33 QDP
        ↳34 UsableRulesProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 QDPSizeChangeProof (⇔, 0 ms)
        ↳37 YES
    ↳38 QDP
        ↳39 UsableRulesProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 QDPOrderProof (⇔, 9142 ms)
        ↳42 QDP
        ↳43 UsableRulesProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 QDPOrderProof (⇔, 3195 ms)
        ↳46 QDP
        ↳47 UsableRulesProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 QDPOrderProof (⇔, 1883 ms)
        ↳50 QDP
        ↳51 DependencyGraphProof (⇔, 0 ms)
        ↳52 TRUE

(0) Obligation:

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

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(d(x)) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right3(c(End(x))) → Left(a(b(c(a(End(x))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
c(d(x)) → a(b(c(a(x))))
a(a(x)) → a(c(b(a(x))))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Ac(x1)) = x1   
POL(Ad(x1)) = 2 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = 2 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Begin(d(x)) → Wait(Right3(x))
Right3(c(End(x))) → Left(a(b(c(a(End(x))))))
c(d(x)) → a(b(c(a(x))))


(2) Obligation:

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

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

BEGIN(c(a(x))) → WAIT(Right1(x))
BEGIN(c(a(x))) → RIGHT1(x)
BEGIN(a(x)) → WAIT(Right2(x))
BEGIN(a(x)) → RIGHT2(x)
BEGIN(a(x)) → WAIT(Right4(x))
BEGIN(a(x)) → RIGHT4(x)
RIGHT1(b(End(x))) → A(b(a(b(c(End(x))))))
RIGHT1(b(End(x))) → B(a(b(c(End(x)))))
RIGHT1(b(End(x))) → A(b(c(End(x))))
RIGHT1(b(End(x))) → B(c(End(x)))
RIGHT2(b(c(End(x)))) → A(b(a(b(c(End(x))))))
RIGHT2(b(c(End(x)))) → B(a(b(c(End(x)))))
RIGHT2(b(c(End(x)))) → A(b(c(End(x))))
RIGHT4(a(End(x))) → A(c(b(a(End(x)))))
RIGHT4(a(End(x))) → B(a(End(x)))
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT4(b(x)) → AB(Right4(x))
RIGHT4(b(x)) → RIGHT4(x)
RIGHT1(c(x)) → AC(Right1(x))
RIGHT1(c(x)) → RIGHT1(x)
RIGHT2(c(x)) → AC(Right2(x))
RIGHT2(c(x)) → RIGHT2(x)
RIGHT3(c(x)) → AC(Right3(x))
RIGHT3(c(x)) → RIGHT3(x)
RIGHT4(c(x)) → AC(Right4(x))
RIGHT4(c(x)) → RIGHT4(x)
RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
RIGHT4(a(x)) → AA(Right4(x))
RIGHT4(a(x)) → RIGHT4(x)
RIGHT1(d(x)) → AD(Right1(x))
RIGHT1(d(x)) → RIGHT1(x)
RIGHT2(d(x)) → AD(Right2(x))
RIGHT2(d(x)) → RIGHT2(x)
RIGHT3(d(x)) → AD(Right3(x))
RIGHT3(d(x)) → RIGHT3(x)
RIGHT4(d(x)) → AD(Right4(x))
RIGHT4(d(x)) → RIGHT4(x)
AB(Left(x)) → B(x)
AA(Left(x)) → A(x)
WAIT(Left(x)) → BEGIN(x)
B(c(a(x))) → A(b(a(b(c(x)))))
B(c(a(x))) → B(a(b(c(x))))
B(c(a(x))) → A(b(c(x)))
B(c(a(x))) → B(c(x))
A(a(x)) → A(c(b(a(x))))
A(a(x)) → B(a(x))

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 31 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

A(a(x)) → B(a(x))
B(c(a(x))) → A(b(a(b(c(x)))))
B(c(a(x))) → B(a(b(c(x))))
B(c(a(x))) → A(b(c(x)))
B(c(a(x))) → B(c(x))

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(8) 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.

(9) Obligation:

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

A(a(x)) → B(a(x))
B(c(a(x))) → A(b(a(b(c(x)))))
B(c(a(x))) → B(a(b(c(x))))
B(c(a(x))) → A(b(c(x)))
B(c(a(x))) → B(c(x))

The TRS R consists of the following rules:

b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


A(a(x)) → B(a(x))
B(c(a(x))) → A(b(a(b(c(x)))))
B(c(a(x))) → B(a(b(c(x))))
B(c(a(x))) → A(b(c(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( A(x1) ) = 1

POL( B(x1) ) = max{0, 2x1 - 2}

POL( a(x1) ) = max{0, -2}

POL( c(x1) ) = 2

POL( b(x1) ) = max{0, -2}


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

a(a(x)) → a(c(b(a(x))))

(11) Obligation:

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

B(c(a(x))) → B(c(x))

The TRS R consists of the following rules:

b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(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:

B(c(a(x))) → B(c(x))

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


B(c(a(x))) → B(c(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(B(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(c(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(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) YES

(18) Obligation:

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

RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(19) 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.

(20) Obligation:

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

RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)

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

(21) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT3(c(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(b(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(a(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(d(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

(22) YES

(23) Obligation:

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

RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(24) 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.

(25) Obligation:

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

RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT4(c(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(b(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(a(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(d(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

(27) YES

(28) Obligation:

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

RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(29) 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.

(30) Obligation:

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

RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)

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

(31) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT2(c(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(b(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(a(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(d(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

(32) YES

(33) Obligation:

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

RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))

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

(34) 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.

(35) Obligation:

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

RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)

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

(36) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT1(c(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(b(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(a(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(d(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

(37) YES

(38) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(c(a(x))) → WAIT(Right1(x))
BEGIN(a(x)) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Begin(c(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(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:

WAIT(Left(x)) → BEGIN(x)
BEGIN(c(a(x))) → WAIT(Right1(x))
BEGIN(a(x)) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(d(x)) → Ad(Right1(x))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BEGIN(c(a(x))) → WAIT(Right1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^5, +, *, >=, >) :

POL(WAIT(x1)) = 0 +
[0,0,1,0,0]
·x1

POL(Left(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00001\
|00000|
|01000|
|00000|
\00010/
·x1

POL(BEGIN(x1)) = 0 +
[0,1,0,0,0]
·x1

POL(c(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00000\
|00001|
|00000|
|00001|
\00000/
·x1

POL(a(x1)) =
/0\
|0|
|0|
|0|
\1/
 +
/00000\
|00010|
|00000|
|00010|
\00010/
·x1

POL(Right1(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00001\
|10000|
|00010|
|00000|
\00010/
·x1

POL(Right2(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00001\
|00000|
|00010|
|00000|
\00010/
·x1

POL(Right4(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00001\
|00000|
|00010|
|00000|
\00010/
·x1

POL(b(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00010\
|00000|
|00000|
|00000|
\00010/
·x1

POL(End(x1)) =
/0\
|0|
|0|
|1|
\1/
 +
/00000\
|00000|
|00000|
|00000|
\01000/
·x1

POL(Ab(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00001\
|00000|
|00000|
|00000|
\00000/
·x1

POL(Ac(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00000\
|00000|
|10000|
|00000|
\10000/
·x1

POL(Aa(x1)) =
/1\
|0|
|0|
|0|
\0/
 +
/00001\
|00000|
|00001|
|00000|
\00001/
·x1

POL(d(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00000\
|00000|
|00000|
|00000|
\00000/
·x1

POL(Ad(x1)) =
/0\
|0|
|0|
|0|
\0/
 +
/00000\
|00000|
|00000|
|00000|
\00000/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(d(x)) → Ad(Right1(x))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))

(42) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(x)) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))
Right1(b(End(x))) → Left(a(b(a(b(c(End(x)))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(d(x)) → Ad(Right1(x))

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

(43) 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.

(44) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(x)) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BEGIN(a(x)) → WAIT(Right2(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(WAIT(x1)) = -I +
[0A,0A,-I]
·x1

POL(Left(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A-I\
|0A-I-I|
\-I-I0A/
·x1

POL(BEGIN(x1)) = -I +
[0A,0A,-I]
·x1

POL(a(x1)) =
/0A\
|1A|
\0A/
 +
/-I-I0A\
|1A1A0A|
\0A0A0A/
·x1

POL(Right2(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A-I\
|0A-I-I|
\-I-I0A/
·x1

POL(Right4(x1)) =
/0A\
|0A|
\0A/
 +
/1A1A0A\
|1A-I-I|
\-I-I1A/
·x1

POL(b(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|0A-I-I|
\-I-I-I/
·x1

POL(c(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I1A\
|-I-I0A|
\-I-I-I/
·x1

POL(End(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|-I0A0A|
\0A0A0A/
·x1

POL(Ab(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|-I-I-I|
\-I-I-I/
·x1

POL(Ac(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I1A\
|-I-I1A|
\-I-I-I/
·x1

POL(Aa(x1)) =
/1A\
|0A|
\0A/
 +
/1A-I0A\
|-I-I0A|
\0A0A0A/
·x1

POL(d(x1)) =
/-I\
|-I|
\0A/
 +
/0A-I-I\
|0A0A0A|
\0A-I0A/
·x1

POL(Ad(x1)) =
/-I\
|-I|
\0A/
 +
/0A0A0A\
|-I0A-I|
\-I0A0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))
Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))

(46) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Right2(b(c(End(x)))) → Left(a(b(a(b(c(End(x)))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))

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

(47) 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.

(48) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


WAIT(Left(x)) → BEGIN(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(WAIT(x1)) = -I +
[0A,-I,-I]
·x1

POL(Left(x1)) =
/1A\
|0A|
\0A/
 +
/1A-I-I\
|-I1A-I|
\-I-I1A/
·x1

POL(BEGIN(x1)) = 0A +
[0A,-I,-I]
·x1

POL(a(x1)) =
/1A\
|-I|
\0A/
 +
/1A0A0A\
|-I-I-I|
\0A-I0A/
·x1

POL(Right4(x1)) =
/0A\
|0A|
\0A/
 +
/1A0A0A\
|-I1A0A|
\0A0A1A/
·x1

POL(End(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|0A0A0A|
\-I0A-I/
·x1

POL(c(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|0A0A0A|
\0A-I-I/
·x1

POL(b(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I0A\
|0A0A0A|
\0A-I0A/
·x1

POL(Ab(x1)) =
/1A\
|-I|
\-I/
 +
/-I-I0A\
|0A0A0A|
\0A-I0A/
·x1

POL(Ac(x1)) =
/1A\
|-I|
\-I/
 +
/-I-I-I\
|0A0A0A|
\0A-I-I/
·x1

POL(Aa(x1)) =
/-I\
|0A|
\1A/
 +
/1A0A0A\
|-I-I-I|
\0A-I0A/
·x1

POL(d(x1)) =
/0A\
|-I|
\0A/
 +
/-I0A0A\
|-I-I-I|
\0A-I-I/
·x1

POL(Ad(x1)) =
/1A\
|0A|
\-I/
 +
/-I0A0A\
|-I-I-I|
\0A-I-I/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
b(x) → c(c(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))

(50) Obligation:

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

BEGIN(a(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(a(End(x))) → Left(a(c(b(a(End(x))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → a(c(b(a(x))))
b(c(a(x))) → a(b(a(b(c(x)))))
b(x) → c(c(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))

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

(51) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(52) TRUE