NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z046-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 9 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 7 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 MNOCProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MNOCProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 NonTerminationLoopProof (⇒, 10 ms)
        ↳13 NO
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 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

(0) Obligation:

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

Begin(a(a(b(a(x))))) → Wait(Right1(x))
Begin(a(b(a(x)))) → Wait(Right2(x))
Begin(b(a(x))) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(c(End(x))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right2(c(a(End(x)))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right3(c(a(a(End(x))))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right4(c(a(a(b(End(x)))))) → Left(a(a(b(a(c(c(a(a(b(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(a(a(b(a(x))))) → a(a(b(a(c(c(a(a(b(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(b(a(x))))) → WAIT(Right1(x))
BEGIN(a(a(b(a(x))))) → RIGHT1(x)
BEGIN(a(b(a(x)))) → WAIT(Right2(x))
BEGIN(a(b(a(x)))) → RIGHT2(x)
BEGIN(b(a(x))) → WAIT(Right3(x))
BEGIN(b(a(x))) → RIGHT3(x)
BEGIN(a(x)) → WAIT(Right4(x))
BEGIN(a(x)) → RIGHT4(x)
RIGHT1(c(End(x))) → C(c(a(a(b(End(x))))))
RIGHT1(c(End(x))) → C(a(a(b(End(x)))))
RIGHT2(c(a(End(x)))) → C(c(a(a(b(End(x))))))
RIGHT2(c(a(End(x)))) → C(a(a(b(End(x)))))
RIGHT3(c(a(a(End(x))))) → C(c(a(a(b(End(x))))))
RIGHT3(c(a(a(End(x))))) → C(a(a(b(End(x)))))
RIGHT4(c(a(a(b(End(x)))))) → C(c(a(a(b(End(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(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)
AC(Left(x)) → C(x)
WAIT(Left(x)) → BEGIN(x)
C(a(a(b(a(x))))) → C(c(a(a(b(x)))))
C(a(a(b(a(x))))) → C(a(a(b(x))))

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

(8) MNOCProof (EQUIVALENT transformation)

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

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

c(a(a(b(a(x0)))))

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

(10) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(11) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(12) 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 = C(c(a(a(b(c(c(a(a(b(a(x'))))))))))) evaluates to t =C(c(a(a(b(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x')))))))))))))))))))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x' / b(a(c(c(a(a(b(c(c(a(a(b(x'))))))))))))]
  • Semiunifier: [ ]



Rewriting sequence

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

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

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

C(a(a(b(a(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x'))))))))))))))))))))))) → C(c(a(a(b(c(c(a(a(b(a(b(a(c(c(a(a(b(c(c(a(a(b(x')))))))))))))))))))))))
with rule C(a(a(b(a(x))))) → C(c(a(a(b(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.



(13) NO

(14) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(b(a(x))))) → Wait(Right1(x))
Begin(a(b(a(x)))) → Wait(Right2(x))
Begin(b(a(x))) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(c(End(x))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right2(c(a(End(x)))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right3(c(a(a(End(x))))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right4(c(a(a(b(End(x)))))) → Left(a(a(b(a(c(c(a(a(b(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(a(a(b(a(x))))) → a(a(b(a(c(c(a(a(b(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(c(x)) → RIGHT4(x)
RIGHT4(b(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(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

(18) YES

(19) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(b(a(x))))) → Wait(Right1(x))
Begin(a(b(a(x)))) → Wait(Right2(x))
Begin(b(a(x))) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(c(End(x))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right2(c(a(End(x)))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right3(c(a(a(End(x))))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right4(c(a(a(b(End(x)))))) → Left(a(a(b(a(c(c(a(a(b(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(a(a(b(a(x))))) → a(a(b(a(c(c(a(a(b(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(c(x)) → RIGHT3(x)
RIGHT3(b(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(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

(23) YES

(24) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(b(a(x))))) → Wait(Right1(x))
Begin(a(b(a(x)))) → Wait(Right2(x))
Begin(b(a(x))) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(c(End(x))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right2(c(a(End(x)))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right3(c(a(a(End(x))))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right4(c(a(a(b(End(x)))))) → Left(a(a(b(a(c(c(a(a(b(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(a(a(b(a(x))))) → a(a(b(a(c(c(a(a(b(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(c(x)) → RIGHT2(x)
RIGHT2(b(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(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

(28) YES

(29) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(b(a(x))))) → Wait(Right1(x))
Begin(a(b(a(x)))) → Wait(Right2(x))
Begin(b(a(x))) → Wait(Right3(x))
Begin(a(x)) → Wait(Right4(x))
Right1(c(End(x))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right2(c(a(End(x)))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right3(c(a(a(End(x))))) → Left(a(a(b(a(c(c(a(a(b(End(x)))))))))))
Right4(c(a(a(b(End(x)))))) → Left(a(a(b(a(c(c(a(a(b(End(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(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(a(a(b(a(x))))) → a(a(b(a(c(c(a(a(b(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(c(x)) → RIGHT1(x)
RIGHT1(b(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(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

(33) YES

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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