YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z064-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 3 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 12 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 187 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 2 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔, 0 ms)
        ↳33 YES
    ↳34 QDP
        ↳35 UsableRulesProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 QDPOrderProof (⇔, 2255 ms)
        ↳38 QDP
        ↳39 UsableRulesProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 QDPOrderProof (⇔, 1691 ms)
        ↳42 QDP
        ↳43 UsableRulesProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 QDPOrderProof (⇔, 14 ms)
        ↳46 QDP
        ↳47 DependencyGraphProof (⇔, 0 ms)
        ↳48 TRUE

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

Begin(a(c(b(b(x))))) → Wait(Right1(x))
Begin(c(b(b(x)))) → Wait(Right2(x))
Begin(b(b(x))) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right2(b(a(End(x)))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(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(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(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))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(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 6 SCCs with 28 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

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

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

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

(9) Obligation:

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

B(a(c(b(b(x))))) → B(b(b(a(c(x)))))

The TRS R consists of the following rules:

b(a(c(b(b(x))))) → a(c(b(b(b(a(c(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.


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

POL(B(x1)) = 0A +
[-I,0A,0A]
·x1

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

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

POL(b(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A1A\
|0A0A0A|
\0A0A0A/
·x1

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

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

(11) Obligation:

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

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

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) YES

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) 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.

(16) Obligation:

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

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

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

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

(18) YES

(19) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(20) 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.

(21) Obligation:

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

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

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

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

(23) YES

(24) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(25) 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.

(26) Obligation:

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

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

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

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

(28) YES

(29) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

(33) YES

(34) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(c(b(b(x))))) → WAIT(Right1(x))
BEGIN(c(b(b(x)))) → WAIT(Right2(x))
BEGIN(b(b(x))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

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

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

(35) 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.

(36) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(c(b(b(x))))) → WAIT(Right1(x))
BEGIN(c(b(b(x)))) → WAIT(Right2(x))
BEGIN(b(b(x))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

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

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

(37) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(c(b(b(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)) = 0A +
[0A,0A,0A]
·x1

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

POL(BEGIN(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

POL(b(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A0A\
|-I0A0A|
\-I0A0A/
·x1

POL(Right1(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A0A\
|-I-I0A|
\-I-I0A/
·x1

POL(Right2(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A0A\
|-I-I0A|
\-I-I0A/
·x1

POL(Right3(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A0A\
|0A-I-I|
\-I-I0A/
·x1

POL(Right4(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A0A\
|0A-I0A|
\-I-I0A/
·x1

POL(End(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A-I\
|0A0A-I|
\0A0A-I/
·x1

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

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

POL(Ac(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A1A\
|0A0A0A|
\0A0A0A/
·x1

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

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

(38) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(c(b(b(x))))) → WAIT(Right1(x))
BEGIN(b(b(x))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right2(b(a(End(x)))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right2(b(x)) → Ab(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right1(b(End(x))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(c(x)) → Ac(Right1(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(a(c(b(b(x))))) → WAIT(Right1(x))
BEGIN(b(b(x))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right1(b(End(x))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(c(x)) → Ac(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(a(c(b(b(x))))) → WAIT(Right1(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,0A]
·x1

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

POL(BEGIN(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

POL(b(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A1A\
|0A0A0A|
\0A-I0A/
·x1

POL(Right1(x1)) =
/-I\
|-I|
\-I/
 +
/0A-I-I\
|-I-I0A|
\0A-I0A/
·x1

POL(Right3(x1)) =
/0A\
|-I|
\0A/
 +
/1A-I-I\
|0A0A1A|
\1A0A0A/
·x1

POL(Right4(x1)) =
/-I\
|-I|
\-I/
 +
/0A-I-I\
|-I-I0A|
\0A-I0A/
·x1

POL(End(x1)) =
/-I\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A1A/
·x1

POL(Ab(x1)) =
/-I\
|-I|
\-I/
 +
/0A1A0A\
|0A0A0A|
\0A1A0A/
·x1

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

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

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

Right1(b(End(x))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))

(42) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(x))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right1(b(End(x))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(c(x)) → Ac(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(b(b(x))) → WAIT(Right3(x))
BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(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.


WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(x))) → WAIT(Right3(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = 1   
POL(Ab(x1)) = 1 + x1   
POL(Ac(x1)) = 1   
POL(BEGIN(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = 1 + x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = 1 + x1   
POL(WAIT(x1)) = x1   
POL(a(x1)) = 0   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 0   

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

Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))

(46) Obligation:

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

BEGIN(b(x)) → WAIT(Right4(x))

The TRS R consists of the following rules:

Right4(b(a(c(b(End(x)))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
b(a(c(b(b(x))))) → a(c(b(b(b(a(c(x)))))))
Right3(b(a(c(End(x))))) → Left(a(c(b(b(b(a(c(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))

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

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(48) TRUE