NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z128.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 38 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 52 ms)
        ↳9 QDP
        ↳10 NonTerminationLoopProof (⇒, 59 ms)
        ↳11 NO
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES
    ↳17 QDP
        ↳18 UsableRulesProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 UsableRulesProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QDPSizeChangeProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP

(0) Obligation:

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

Begin(a(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(b(x))) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Right1(a(End(x))) → Left(b(a(b(End(x)))))
Right2(a(a(End(x)))) → Left(b(a(b(End(x)))))
Right3(b(End(x))) → Left(a(b(a(End(x)))))
Right4(b(a(End(x)))) → Left(a(b(a(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(a(a(x))) → b(a(b(x)))
b(a(b(x))) → a(b(a(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(a(x))) → WAIT(Right1(x))
BEGIN(a(a(x))) → RIGHT1(x)
BEGIN(a(x)) → WAIT(Right2(x))
BEGIN(a(x)) → RIGHT2(x)
BEGIN(a(b(x))) → WAIT(Right3(x))
BEGIN(a(b(x))) → RIGHT3(x)
BEGIN(b(x)) → WAIT(Right4(x))
BEGIN(b(x)) → RIGHT4(x)
RIGHT1(a(End(x))) → B(a(b(End(x))))
RIGHT1(a(End(x))) → A(b(End(x)))
RIGHT1(a(End(x))) → B(End(x))
RIGHT2(a(a(End(x)))) → B(a(b(End(x))))
RIGHT2(a(a(End(x)))) → A(b(End(x)))
RIGHT2(a(a(End(x)))) → B(End(x))
RIGHT3(b(End(x))) → A(b(a(End(x))))
RIGHT3(b(End(x))) → B(a(End(x)))
RIGHT3(b(End(x))) → A(End(x))
RIGHT4(b(a(End(x)))) → A(b(a(End(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(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)
AA(Left(x)) → A(x)
AB(Left(x)) → B(x)
WAIT(Left(x)) → BEGIN(x)
A(a(a(x))) → B(a(b(x)))
A(a(a(x))) → A(b(x))
A(a(a(x))) → B(x)
B(a(b(x))) → A(b(a(x)))
B(a(b(x))) → B(a(x))
B(a(b(x))) → A(x)

The TRS R consists of the following rules:

Begin(a(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(b(x))) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Right1(a(End(x))) → Left(b(a(b(End(x)))))
Right2(a(a(End(x)))) → Left(b(a(b(End(x)))))
Right3(b(End(x))) → Left(a(b(a(End(x)))))
Right4(b(a(End(x)))) → Left(a(b(a(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(a(a(x))) → b(a(b(x)))
b(a(b(x))) → a(b(a(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 24 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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

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

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = A(b(a(b(a(a(x)))))) evaluates to t =A(b(a(b(a(a(x))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

A(b(a(b(a(a(x)))))) → A(a(b(a(a(a(x))))))
with rule b(a(b(x'))) → a(b(a(x'))) at position [0] and matcher [x' / a(a(x))]

A(a(b(a(a(a(x)))))) → A(a(b(b(a(b(x))))))
with rule a(a(a(x'))) → b(a(b(x'))) at position [0,0,0] and matcher [x' / x]

A(a(b(b(a(b(x)))))) → A(a(b(a(b(a(x))))))
with rule b(a(b(x'))) → a(b(a(x'))) at position [0,0,0] and matcher [x' / x]

A(a(b(a(b(a(x)))))) → A(a(a(b(a(a(x))))))
with rule b(a(b(x'))) → a(b(a(x'))) at position [0,0] and matcher [x' / a(x)]

A(a(a(b(a(a(x)))))) → B(a(b(b(a(a(x))))))
with rule A(a(a(x'))) → B(a(b(x'))) at position [] and matcher [x' / b(a(a(x)))]

B(a(b(b(a(a(x)))))) → A(b(a(b(a(a(x))))))
with rule B(a(b(x))) → A(b(a(x)))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(11) NO

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

(15) 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(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

(16) YES

(17) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(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:

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

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

(20) 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(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

(21) YES

(22) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(23) 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.

(24) Obligation:

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

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

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

(25) 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(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

(26) YES

(27) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(28) 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.

(29) Obligation:

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

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

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

(30) 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(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

(31) YES

(32) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(33) 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.

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

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