YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Waldmann_06_SRS/e.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 15 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 QDPOrderProof (⇔, 65 ms)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔, 368 ms)
        ↳14 QDP
        ↳15 DependencyGraphProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPOrderProof (⇔, 131 ms)
        ↳18 QDP
        ↳19 QDPOrderProof (⇔, 283 ms)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 256 ms)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPOrderProof (⇔, 217 ms)
        ↳28 QDP
        ↳29 QDPOrderProof (⇔, 7 ms)
        ↳30 QDP
        ↳31 DependencyGraphProof (⇔, 0 ms)
        ↳32 TRUE

(0) Obligation:

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

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(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:

21(8(x)) → 41(x)
51(9(x)) → 01(x)
41(x) → 51(2(3(x)))
41(x) → 21(3(x))
51(3(x)) → 61(0(x))
51(3(x)) → 01(x)
21(8(x)) → 71(x)
51(2(6(x))) → 61(2(4(x)))
51(2(6(x))) → 21(4(x))
51(2(6(x))) → 41(x)
91(7(x)) → 71(5(x))
91(7(x)) → 51(x)
71(2(x)) → 41(x)
71(0(x)) → 91(3(x))
91(5(9(x))) → 51(7(x))
91(5(9(x))) → 71(x)
41(x) → 91(6(6(x)))
41(x) → 61(6(x))
41(x) → 61(x)
91(x) → 61(7(x))
91(x) → 71(x)
61(2(x)) → 71(7(x))
61(2(x)) → 71(x)
21(4(x)) → 01(7(x))
21(4(x)) → 71(x)
01(3(x)) → 51(3(x))

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(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 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

51(3(x)) → 01(x)
01(3(x)) → 51(3(x))

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(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:

51(3(x)) → 01(x)
01(3(x)) → 51(3(x))

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

(8) 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:

  • 01(3(x)) → 51(3(x))
    The graph contains the following edges 1 >= 1

  • 51(3(x)) → 01(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

41(x) → 91(6(6(x)))
91(7(x)) → 71(5(x))
71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 61(x)
61(2(x)) → 71(7(x))
61(2(x)) → 71(x)
91(7(x)) → 51(x)
51(2(6(x))) → 61(2(4(x)))
51(2(6(x))) → 21(4(x))
21(8(x)) → 41(x)
21(8(x)) → 71(x)
21(4(x)) → 71(x)
51(2(6(x))) → 41(x)
91(5(9(x))) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(11) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


21(8(x)) → 41(x)
21(8(x)) → 71(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 0   
POL(1(x1)) = 0   
POL(2(x1)) = 0   
POL(21(x1)) = x1   
POL(3(x1)) = 0   
POL(4(x1)) = 0   
POL(41(x1)) = 0   
POL(5(x1)) = 0   
POL(51(x1)) = 0   
POL(6(x1)) = 0   
POL(61(x1)) = 0   
POL(7(x1)) = 0   
POL(71(x1)) = 0   
POL(8(x1)) = 1   
POL(9(x1)) = 0   
POL(91(x1)) = 0   

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))

(12) Obligation:

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

41(x) → 91(6(6(x)))
91(7(x)) → 71(5(x))
71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 61(x)
61(2(x)) → 71(7(x))
61(2(x)) → 71(x)
91(7(x)) → 51(x)
51(2(6(x))) → 61(2(4(x)))
51(2(6(x))) → 21(4(x))
21(4(x)) → 71(x)
51(2(6(x))) → 41(x)
91(5(9(x))) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(13) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


91(7(x)) → 51(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(41(x1)) = 1A +
[1A,1A,1A]
·x1

POL(91(x1)) = 1A +
[0A,1A,-I]
·x1

POL(6(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A-I0A/
·x1

POL(7(x1)) =
/-I\
|-I|
\0A/
 +
/0A-I0A\
|0A0A0A|
\0A0A0A/
·x1

POL(71(x1)) = 1A +
[0A,1A,-I]
·x1

POL(5(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

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

POL(3(x1)) =
/0A\
|-I|
\0A/
 +
/0A-I-I\
|-I-I-I|
\0A0A0A/
·x1

POL(2(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A1A0A|
\0A0A0A/
·x1

POL(61(x1)) = 0A +
[0A,0A,1A]
·x1

POL(51(x1)) = 0A +
[0A,0A,0A]
·x1

POL(4(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(21(x1)) = 1A +
[-I,1A,-I]
·x1

POL(9(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

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

POL(8(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I-I|
\0A0A0A/
·x1

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))
2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
2(8(x)) → 7(x)
2(4(x)) → 0(7(x))

(14) Obligation:

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

41(x) → 91(6(6(x)))
91(7(x)) → 71(5(x))
71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 61(x)
61(2(x)) → 71(7(x))
61(2(x)) → 71(x)
51(2(6(x))) → 61(2(4(x)))
51(2(6(x))) → 21(4(x))
21(4(x)) → 71(x)
51(2(6(x))) → 41(x)
91(5(9(x))) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(16) Obligation:

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

91(7(x)) → 71(5(x))
71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(5(9(x))) → 71(x)
41(x) → 61(x)
61(2(x)) → 71(7(x))
61(2(x)) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(17) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


61(2(x)) → 71(7(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(91(x1)) = 0A +
[0A,0A,-I]
·x1

POL(7(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I0A\
|0A0A0A|
\-I-I0A/
·x1

POL(71(x1)) = 0A +
[0A,-I,-I]
·x1

POL(5(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I0A\
|-I-I0A|
\-I-I0A/
·x1

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

POL(3(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1

POL(2(x1)) =
/0A\
|1A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A1A1A/
·x1

POL(41(x1)) = 0A +
[0A,0A,0A]
·x1

POL(6(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A-I\
|0A-I0A|
\-I0A0A/
·x1

POL(9(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|-I-I0A|
\0A0A0A/
·x1

POL(61(x1)) = 0A +
[0A,0A,0A]
·x1

POL(4(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A1A\
|0A0A0A|
\0A0A0A/
·x1

POL(8(x1)) =
/0A\
|-I|
\0A/
 +
/1A1A1A\
|0A0A0A|
\0A-I0A/
·x1

POL(1(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|-I0A0A|
\0A-I-I/
·x1

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
2(8(x)) → 4(x)
2(8(x)) → 7(x)
2(4(x)) → 0(7(x))
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))
2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)

(18) Obligation:

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

91(7(x)) → 71(5(x))
71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(5(9(x))) → 71(x)
41(x) → 61(x)
61(2(x)) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(19) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


61(2(x)) → 71(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(91(x1)) = 0A +
[0A,0A,0A]
·x1

POL(7(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I-I\
|0A-I0A|
\0A0A-I/
·x1

POL(71(x1)) = 0A +
[0A,0A,0A]
·x1

POL(5(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I0A\
|0A0A0A|
\0A-I0A/
·x1

POL(0(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A-I\
|0A0A-I|
\0A-I-I/
·x1

POL(3(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A-I\
|-I-I-I|
\-I-I-I/
·x1

POL(2(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A0A\
|0A0A0A|
\0A1A1A/
·x1

POL(41(x1)) = 0A +
[0A,1A,1A]
·x1

POL(6(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A-I0A|
\0A0A0A/
·x1

POL(9(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A-I|
\0A0A0A/
·x1

POL(61(x1)) = 0A +
[0A,1A,0A]
·x1

POL(4(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(8(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(1(x1)) =
/0A\
|-I|
\0A/
 +
/0A-I-I\
|-I0A0A|
\0A0A0A/
·x1

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
2(8(x)) → 4(x)
2(8(x)) → 7(x)
2(4(x)) → 0(7(x))
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))
2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)

(20) Obligation:

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

91(7(x)) → 71(5(x))
71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(5(9(x))) → 71(x)
41(x) → 61(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

71(0(x)) → 91(3(x))
91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(7(x)) → 71(5(x))
91(5(9(x))) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(23) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


71(0(x)) → 91(3(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(71(x1)) = 0A +
[0A,-I,0A]
·x1

POL(0(x1)) =
/1A\
|1A|
\0A/
 +
/-I-I-I\
|-I-I-I|
\0A0A-I/
·x1

POL(91(x1)) = 0A +
[0A,-I,0A]
·x1

POL(3(x1)) =
/0A\
|1A|
\0A/
 +
/-I-I-I\
|0A0A-I|
\-I-I-I/
·x1

POL(2(x1)) =
/0A\
|1A|
\1A/
 +
/0A0A-I\
|0A0A-I|
\0A0A0A/
·x1

POL(41(x1)) = 1A +
[0A,0A,0A]
·x1

POL(6(x1)) =
/0A\
|1A|
\0A/
 +
/0A0A-I\
|0A-I-I|
\0A0A0A/
·x1

POL(7(x1)) =
/1A\
|1A|
\0A/
 +
/-I-I-I\
|-I-I-I|
\0A0A0A/
·x1

POL(5(x1)) =
/1A\
|1A|
\0A/
 +
/-I-I-I\
|-I-I-I|
\0A0A0A/
·x1

POL(9(x1)) =
/1A\
|1A|
\1A/
 +
/-I-I-I\
|-I-I-I|
\0A0A0A/
·x1

POL(4(x1)) =
/1A\
|1A|
\1A/
 +
/-I-I-I\
|-I-I-I|
\0A0A0A/
·x1

POL(8(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

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

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
2(8(x)) → 4(x)
2(8(x)) → 7(x)
2(4(x)) → 0(7(x))
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))
2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)

(24) Obligation:

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

91(x) → 71(x)
71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(7(x)) → 71(5(x))
91(5(9(x))) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) Obligation:

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

71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(5(9(x))) → 71(x)
91(x) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(27) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


91(5(9(x))) → 71(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(71(x1)) = 0A +
[-I,0A,-I]
·x1

POL(2(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A1A\
|1A1A1A|
\0A0A0A/
·x1

POL(41(x1)) = 0A +
[1A,1A,1A]
·x1

POL(91(x1)) = 0A +
[-I,1A,-I]
·x1

POL(6(x1)) =
/-I\
|-I|
\0A/
 +
/0A-I0A\
|0A0A-I|
\0A0A0A/
·x1

POL(5(x1)) =
/-I\
|0A|
\-I/
 +
/0A-I-I\
|0A-I0A|
\0A-I0A/
·x1

POL(9(x1)) =
/-I\
|0A|
\0A/
 +
/0A0A0A\
|0A-I0A|
\0A0A0A/
·x1

POL(0(x1)) =
/-I\
|0A|
\0A/
 +
/0A0A-I\
|0A0A-I|
\-I0A-I/
·x1

POL(3(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A-I\
|0A0A-I|
\-I-I-I/
·x1

POL(7(x1)) =
/-I\
|0A|
\-I/
 +
/0A-I0A\
|0A-I0A|
\0A0A0A/
·x1

POL(4(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(8(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(1(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A0A\
|0A0A0A|
\0A0A-I/
·x1

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
2(8(x)) → 4(x)
2(8(x)) → 7(x)
2(4(x)) → 0(7(x))
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))
2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)

(28) Obligation:

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

71(2(x)) → 41(x)
41(x) → 91(6(6(x)))
91(x) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(29) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


71(2(x)) → 41(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 + x1   
POL(3(x1)) = 0   
POL(4(x1)) = 0   
POL(41(x1)) = 1   
POL(5(x1)) = 0   
POL(6(x1)) = 0   
POL(7(x1)) = 0   
POL(71(x1)) = 1 + x1   
POL(8(x1)) = 1   
POL(9(x1)) = 0   
POL(91(x1)) = 1 + x1   

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

0(3(x)) → 5(3(x))
5(3(x)) → 6(0(x))
6(9(x)) → 9(x)
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
4(x) → 9(6(6(x)))
9(5(9(x))) → 5(7(x))
5(9(x)) → 0(x)
5(2(6(x))) → 6(2(4(x)))
6(2(x)) → 7(7(x))
7(0(x)) → 9(3(x))
9(x) → 6(7(x))
6(6(x)) → 3(x)
4(x) → 5(2(3(x)))
4(7(x)) → 1(3(x))

(30) Obligation:

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

41(x) → 91(6(6(x)))
91(x) → 71(x)

The TRS R consists of the following rules:

2(7(x)) → 1(8(x))
2(8(1(x))) → 8(x)
2(8(x)) → 4(x)
5(9(x)) → 0(x)
4(x) → 5(2(3(x)))
5(3(x)) → 6(0(x))
2(8(x)) → 7(x)
4(7(x)) → 1(3(x))
5(2(6(x))) → 6(2(4(x)))
9(7(x)) → 7(5(x))
7(2(x)) → 4(x)
7(0(x)) → 9(3(x))
6(9(x)) → 9(x)
9(5(9(x))) → 5(7(x))
4(x) → 9(6(6(x)))
9(x) → 6(7(x))
6(2(x)) → 7(7(x))
2(4(x)) → 0(7(x))
6(6(x)) → 3(x)
0(3(x)) → 5(3(x))

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

(31) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(32) TRUE