NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z120.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 17 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 20 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 125 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 TRUE
    ↳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
        ↳37 QDPSizeChangeProof (⇔, 0 ms)
        ↳38 YES
    ↳39 QDP
        ↳40 UsableRulesProof (⇔, 7 ms)
        ↳41 QDP
        ↳42 NonTerminationLoopProof (⇒, 569 ms)
        ↳43 NO

(0) Obligation:

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

Begin(c(c(a(x)))) → Wait(Right1(x))
Begin(c(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(c(End(x))) → Left(d(d(End(x))))
Right2(c(c(End(x)))) → Left(d(d(End(x))))
Right3(c(c(c(End(x))))) → Left(d(d(End(x))))
Right4(d(End(x))) → Left(c(c(End(x))))
Right5(b(End(x))) → Left(b(a(c(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))
Right5(c(x)) → Ac(Right5(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(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(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))
Right5(b(x)) → Ab(Right5(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
d(x) → b(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(c(c(a(x)))) → WAIT(Right1(x))
BEGIN(c(c(a(x)))) → RIGHT1(x)
BEGIN(c(a(x))) → WAIT(Right2(x))
BEGIN(c(a(x))) → RIGHT2(x)
BEGIN(a(x)) → WAIT(Right3(x))
BEGIN(a(x)) → RIGHT3(x)
BEGIN(b(x)) → WAIT(Right4(x))
BEGIN(b(x)) → RIGHT4(x)
BEGIN(c(x)) → WAIT(Right5(x))
BEGIN(c(x)) → RIGHT5(x)
RIGHT1(c(End(x))) → D(d(End(x)))
RIGHT1(c(End(x))) → D(End(x))
RIGHT2(c(c(End(x)))) → D(d(End(x)))
RIGHT2(c(c(End(x)))) → D(End(x))
RIGHT3(c(c(c(End(x))))) → D(d(End(x)))
RIGHT3(c(c(c(End(x))))) → D(End(x))
RIGHT4(d(End(x))) → C(c(End(x)))
RIGHT4(d(End(x))) → C(End(x))
RIGHT5(b(End(x))) → B(a(c(End(x))))
RIGHT5(b(End(x))) → C(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)
RIGHT5(c(x)) → AC(Right5(x))
RIGHT5(c(x)) → RIGHT5(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(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)
RIGHT5(d(x)) → AD(Right5(x))
RIGHT5(d(x)) → RIGHT5(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)
RIGHT5(b(x)) → AB(Right5(x))
RIGHT5(b(x)) → RIGHT5(x)
AC(Left(x)) → C(x)
AD(Left(x)) → D(x)
AB(Left(x)) → B(x)
WAIT(Left(x)) → BEGIN(x)
C(c(c(a(x)))) → D(d(x))
C(c(c(a(x)))) → D(x)
D(b(x)) → C(c(x))
D(b(x)) → C(x)
B(c(x)) → B(a(c(x)))
D(x) → B(c(x))
D(x) → C(x)

The TRS R consists of the following rules:

Begin(c(c(a(x)))) → Wait(Right1(x))
Begin(c(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(c(End(x))) → Left(d(d(End(x))))
Right2(c(c(End(x)))) → Left(d(d(End(x))))
Right3(c(c(c(End(x))))) → Left(d(d(End(x))))
Right4(d(End(x))) → Left(c(c(End(x))))
Right5(b(End(x))) → Left(b(a(c(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))
Right5(c(x)) → Ac(Right5(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(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(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))
Right5(b(x)) → Ab(Right5(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
b(c(x)) → b(a(c(x)))
d(x) → b(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 7 SCCs with 40 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

POL(C(x1)) = 1 + x1   
POL(D(x1)) = 2 + x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 2 + x1   
POL(d(x1)) = 3 + x1   

(9) Obligation:

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

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

The TRS R consists of the following rules:

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


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

POL(D(x1)) = 1A +
[0A,-I,1A]
·x1

POL(b(x1)) =
/0A\
|1A|
\-I/
 +
/-I0A0A\
|0A-I1A|
\-I0A0A/
·x1

POL(C(x1)) = 0A +
[-I,0A,-I]
·x1

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

POL(a(x1)) =
/1A\
|0A|
\0A/
 +
/1A1A1A\
|0A0A0A|
\-I0A-I/
·x1

POL(d(x1)) =
/0A\
|1A|
\0A/
 +
/0A0A0A\
|1A0A1A|
\0A0A0A/
·x1

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

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

(11) Obligation:

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

C(c(c(a(x)))) → D(d(x))

The TRS R consists of the following rules:

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

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

(12) DependencyGraphProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

RIGHT5(a(x)) → RIGHT5(x)
RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(d(x)) → RIGHT5(x)
RIGHT5(b(x)) → RIGHT5(x)

The TRS R consists of the following rules:

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

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

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

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

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

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

(18) YES

(19) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

(23) YES

(24) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

(28) YES

(29) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

(33) YES

(34) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

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

(38) YES

(39) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(40) 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.

(41) Obligation:

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

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

The TRS R consists of the following rules:

Right5(b(End(x))) → Left(b(a(c(End(x)))))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Right5(d(x)) → Ad(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Ab(Left(x)) → Left(b(x))
b(c(x)) → b(a(c(x)))
d(b(x)) → c(c(x))
c(c(c(a(x)))) → d(d(x))
d(x) → b(c(x))
Ad(Left(x)) → Left(d(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Right4(d(End(x))) → Left(c(c(End(x))))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right4(d(x)) → Ad(Right4(x))
Right4(b(x)) → Ab(Right4(x))
Right3(c(c(c(End(x))))) → Left(d(d(End(x))))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right3(d(x)) → Ad(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right2(c(c(End(x)))) → Left(d(d(End(x))))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right2(d(x)) → Ad(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right1(c(End(x))) → Left(d(d(End(x))))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right1(d(x)) → Ad(Right1(x))
Right1(b(x)) → Ab(Right1(x))

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

(42) 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 = BEGIN(d(d(End(x')))) evaluates to t =BEGIN(d(d(End(x'))))

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



Rewriting sequence

BEGIN(d(d(End(x')))) → BEGIN(d(b(c(End(x')))))
with rule d(x) → b(c(x)) at position [0,0] and matcher [x / End(x')]

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

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

BEGIN(c(c(a(c(End(x')))))) → WAIT(Right1(c(End(x'))))
with rule BEGIN(c(c(a(x)))) → WAIT(Right1(x)) at position [] and matcher [x / c(End(x'))]

WAIT(Right1(c(End(x')))) → WAIT(Left(d(d(End(x')))))
with rule Right1(c(End(x''))) → Left(d(d(End(x'')))) at position [0] and matcher [x'' / x']

WAIT(Left(d(d(End(x'))))) → BEGIN(d(d(End(x'))))
with rule WAIT(Left(x)) → BEGIN(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.



(43) NO