YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/beans1.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 7 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 48 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPOrderProof (⇔, 21 ms)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔, 0 ms)
        ↳18 YES

(0) Obligation:

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

1(2(1(x))) → 2(0(2(x)))
0(2(1(x))) → 1(0(2(x)))
L(2(1(x))) → L(1(0(2(x))))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(0(x))) → 1(0(1(x)))
L(2(0(x))) → L(1(0(1(x))))
0(2(R(x))) → 1(0(1(R(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:

11(2(1(x))) → 01(2(x))
01(2(1(x))) → 11(0(2(x)))
01(2(1(x))) → 01(2(x))
L1(2(1(x))) → L1(1(0(2(x))))
L1(2(1(x))) → 11(0(2(x)))
L1(2(1(x))) → 01(2(x))
11(2(0(x))) → 01(1(x))
11(2(0(x))) → 11(x)
11(2(R(x))) → 01(1(R(x)))
11(2(R(x))) → 11(R(x))
01(2(0(x))) → 11(0(1(x)))
01(2(0(x))) → 01(1(x))
01(2(0(x))) → 11(x)
L1(2(0(x))) → L1(1(0(1(x))))
L1(2(0(x))) → 11(0(1(x)))
L1(2(0(x))) → 01(1(x))
L1(2(0(x))) → 11(x)
01(2(R(x))) → 11(0(1(R(x))))
01(2(R(x))) → 01(1(R(x)))
01(2(R(x))) → 11(R(x))

The TRS R consists of the following rules:

1(2(1(x))) → 2(0(2(x)))
0(2(1(x))) → 1(0(2(x)))
L(2(1(x))) → L(1(0(2(x))))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(0(x))) → 1(0(1(x)))
L(2(0(x))) → L(1(0(1(x))))
0(2(R(x))) → 1(0(1(R(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 2 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

01(2(1(x))) → 11(0(2(x)))
11(2(1(x))) → 01(2(x))
01(2(1(x))) → 01(2(x))
01(2(0(x))) → 11(0(1(x)))
11(2(0(x))) → 01(1(x))
01(2(0(x))) → 01(1(x))
01(2(0(x))) → 11(x)
11(2(0(x))) → 11(x)

The TRS R consists of the following rules:

1(2(1(x))) → 2(0(2(x)))
0(2(1(x))) → 1(0(2(x)))
L(2(1(x))) → L(1(0(2(x))))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(0(x))) → 1(0(1(x)))
L(2(0(x))) → L(1(0(1(x))))
0(2(R(x))) → 1(0(1(R(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:

01(2(1(x))) → 11(0(2(x)))
11(2(1(x))) → 01(2(x))
01(2(1(x))) → 01(2(x))
01(2(0(x))) → 11(0(1(x)))
11(2(0(x))) → 01(1(x))
01(2(0(x))) → 01(1(x))
01(2(0(x))) → 11(x)
11(2(0(x))) → 11(x)

The TRS R consists of the following rules:

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(1(x))) → 1(0(2(x)))
0(2(0(x))) → 1(0(1(x)))
0(2(R(x))) → 1(0(1(R(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:

01(2(1(x))) → 11(0(2(x)))
11(2(1(x))) → 01(2(x))
01(2(1(x))) → 01(2(x))
01(2(0(x))) → 11(0(1(x)))
11(2(0(x))) → 01(1(x))
01(2(0(x))) → 01(1(x))
01(2(0(x))) → 11(x)
11(2(0(x))) → 11(x)


Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(01(x1)) = 2 + 2·x1   
POL(1(x1)) = 1 + x1   
POL(11(x1)) = 1 + 2·x1   
POL(2(x1)) = 2 + x1   
POL(R(x1)) = x1   

(9) Obligation:

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

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(1(x))) → 1(0(2(x)))
0(2(0(x))) → 1(0(1(x)))
0(2(R(x))) → 1(0(1(R(x))))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) YES

(12) Obligation:

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

L1(2(0(x))) → L1(1(0(1(x))))
L1(2(1(x))) → L1(1(0(2(x))))

The TRS R consists of the following rules:

1(2(1(x))) → 2(0(2(x)))
0(2(1(x))) → 1(0(2(x)))
L(2(1(x))) → L(1(0(2(x))))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(0(x))) → 1(0(1(x)))
L(2(0(x))) → L(1(0(1(x))))
0(2(R(x))) → 1(0(1(R(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:

L1(2(0(x))) → L1(1(0(1(x))))
L1(2(1(x))) → L1(1(0(2(x))))

The TRS R consists of the following rules:

0(2(1(x))) → 1(0(2(x)))
0(2(0(x))) → 1(0(1(x)))
0(2(R(x))) → 1(0(1(R(x))))
1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))

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

(15) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


L1(2(0(x))) → L1(1(0(1(x))))
L1(2(1(x))) → L1(1(0(2(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 0   
POL(1(x1)) = x1   
POL(2(x1)) = 1   
POL(L1(x1)) = x1   
POL(R(x1)) = x1   

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

1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))
0(2(1(x))) → 1(0(2(x)))
0(2(0(x))) → 1(0(1(x)))
0(2(R(x))) → 1(0(1(R(x))))

(16) Obligation:

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

0(2(1(x))) → 1(0(2(x)))
0(2(0(x))) → 1(0(1(x)))
0(2(R(x))) → 1(0(1(R(x))))
1(2(1(x))) → 2(0(2(x)))
1(2(0(x))) → 2(0(1(x)))
1(2(R(x))) → 2(0(1(R(x))))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) YES