NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Waldmann_06_SRS/uni-2.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 24 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 181 ms)
        ↳9 QDP
        ↳10 NonTerminationLoopProof (⇒, 940 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
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP

(0) Obligation:

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

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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(a(x)))) → WAIT(Right1(x))
BEGIN(a(a(a(x)))) → RIGHT1(x)
BEGIN(a(a(x))) → WAIT(Right2(x))
BEGIN(a(a(x))) → RIGHT2(x)
BEGIN(a(x)) → WAIT(Right3(x))
BEGIN(a(x)) → RIGHT3(x)
BEGIN(c(c(x))) → WAIT(Right4(x))
BEGIN(c(c(x))) → RIGHT4(x)
BEGIN(c(x)) → WAIT(Right5(x))
BEGIN(c(x)) → RIGHT5(x)
RIGHT1(a(End(x))) → A(c(a(c(c(End(x))))))
RIGHT1(a(End(x))) → C(a(c(c(End(x)))))
RIGHT1(a(End(x))) → A(c(c(End(x))))
RIGHT1(a(End(x))) → C(c(End(x)))
RIGHT1(a(End(x))) → C(End(x))
RIGHT2(a(a(End(x)))) → A(c(a(c(c(End(x))))))
RIGHT2(a(a(End(x)))) → C(a(c(c(End(x)))))
RIGHT2(a(a(End(x)))) → A(c(c(End(x))))
RIGHT2(a(a(End(x)))) → C(c(End(x)))
RIGHT2(a(a(End(x)))) → C(End(x))
RIGHT3(a(a(a(End(x))))) → A(c(a(c(c(End(x))))))
RIGHT3(a(a(a(End(x))))) → C(a(c(c(End(x)))))
RIGHT3(a(a(a(End(x))))) → A(c(c(End(x))))
RIGHT3(a(a(a(End(x))))) → C(c(End(x)))
RIGHT3(a(a(a(End(x))))) → C(End(x))
RIGHT4(c(End(x))) → A(a(a(End(x))))
RIGHT4(c(End(x))) → A(a(End(x)))
RIGHT4(c(End(x))) → A(End(x))
RIGHT5(c(c(End(x)))) → A(a(a(End(x))))
RIGHT5(c(c(End(x)))) → A(a(End(x)))
RIGHT5(c(c(End(x)))) → 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)
RIGHT5(a(x)) → AA(Right5(x))
RIGHT5(a(x)) → RIGHT5(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)
RIGHT5(c(x)) → AC(Right5(x))
RIGHT5(c(x)) → RIGHT5(x)
AA(Left(x)) → A(x)
AC(Left(x)) → C(x)
WAIT(Left(x)) → BEGIN(x)
A(a(a(a(x)))) → A(c(a(c(c(x)))))
A(a(a(a(x)))) → C(a(c(c(x))))
A(a(a(a(x)))) → A(c(c(x)))
A(a(a(a(x)))) → C(c(x))
A(a(a(a(x)))) → C(x)
C(c(c(x))) → A(a(a(x)))
C(c(c(x))) → A(a(x))
C(c(c(x))) → A(x)

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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 7 SCCs with 39 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A(a(a(a(x)))) → A(c(c(x)))
A(a(a(a(x)))) → A(c(a(c(c(x)))))
A(a(a(a(x)))) → C(c(x))
C(c(c(x))) → A(a(a(x)))
A(a(a(a(x)))) → C(x)
C(c(c(x))) → A(a(x))
C(c(c(x))) → A(x)

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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:

A(a(a(a(x)))) → A(c(c(x)))
A(a(a(a(x)))) → A(c(a(c(c(x)))))
A(a(a(a(x)))) → C(c(x))
C(c(c(x))) → A(a(a(x)))
A(a(a(a(x)))) → C(x)
C(c(c(x))) → A(a(x))
C(c(c(x))) → A(x)

The TRS R consists of the following rules:

a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(a(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.


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

POL(A(x1)) = 0A +
[1A,0A,0A]
·x1

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

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

POL(C(x1)) = 0A +
[0A,0A,1A]
·x1

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

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

(9) Obligation:

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

A(a(a(a(x)))) → A(c(c(x)))
A(a(a(a(x)))) → C(c(x))
C(c(c(x))) → A(a(a(x)))
A(a(a(a(x)))) → C(x)
C(c(c(x))) → A(a(x))
C(c(c(x))) → A(x)

The TRS R consists of the following rules:

a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(a(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(c(c(c(a(c(c(c(a(c(c(c(a(x'))))))))))))) evaluates to t =A(c(c(c(a(c(c(c(a(c(c(c(a(c(c(x')))))))))))))))

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



Rewriting sequence

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

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

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

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

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

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

A(a(a(a(c(a(c(c(c(a(c(c(c(a(c(c(x')))))))))))))))) → A(c(c(c(a(c(c(c(a(c(c(c(a(c(c(x')))))))))))))))
with rule A(a(a(a(x)))) → A(c(c(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:

RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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:

RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(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:

  • RIGHT5(c(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(a(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

(16) YES

(17) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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:

RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(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:

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

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

(21) YES

(22) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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:

RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(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:

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

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

(26) YES

(27) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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:

RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(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:

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

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

(31) YES

(32) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(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:

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

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

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

(36) YES

(37) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(a(a(x)))) → WAIT(Right1(x))
BEGIN(a(a(x))) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right3(x))
BEGIN(c(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right5(c(c(End(x)))) → Left(a(a(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))
Right5(a(x)) → Aa(Right5(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))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → a(c(a(c(c(x)))))
c(c(c(x))) → a(a(a(x)))

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

(38) 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.

(39) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(a(a(a(x)))) → WAIT(Right1(x))
BEGIN(a(a(x))) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right3(x))
BEGIN(c(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))

The TRS R consists of the following rules:

Right5(c(c(End(x)))) → Left(a(a(a(End(x)))))
Right5(a(x)) → Aa(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Ac(Left(x)) → Left(c(x))
c(c(c(x))) → a(a(a(x)))
a(a(a(a(x)))) → a(c(a(c(c(x)))))
Aa(Left(x)) → Left(a(x))
Right4(c(End(x))) → Left(a(a(a(End(x)))))
Right4(a(x)) → Aa(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right3(a(a(a(End(x))))) → Left(a(c(a(c(c(End(x)))))))
Right3(a(x)) → Aa(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right2(a(a(End(x)))) → Left(a(c(a(c(c(End(x)))))))
Right2(a(x)) → Aa(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right1(a(End(x))) → Left(a(c(a(c(c(End(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.