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

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

0 QTRS
↳1 DependencyPairsProof (⇔, 22 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 155 ms)
        ↳9 QDP
        ↳10 DependencyGraphProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 237 ms)
        ↳13 QDP
        ↳14 DependencyGraphProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPOrderProof (⇔, 286 ms)
        ↳17 QDP
        ↳18 DependencyGraphProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPOrderProof (⇔, 9 ms)
        ↳21 QDP
        ↳22 DependencyGraphProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QDPOrderProof (⇔, 15 ms)
        ↳25 QDP
        ↳26 PisEmptyProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QDPSizeChangeProof (⇔, 0 ms)
        ↳32 YES

(0) Obligation:

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

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

51(9(x)) → 21(6(5(x)))
51(9(x)) → 51(x)
31(5(x)) → 81(9(7(x)))
31(5(x)) → 91(7(x))
31(5(x)) → 71(x)
91(x) → 31(2(3(x)))
91(x) → 21(3(x))
91(x) → 31(x)
21(6(x)) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(8(x)) → 71(x)
91(x) → 51(0(2(x)))
91(x) → 21(x)
81(8(4(x))) → 91(x)
71(1(x)) → 91(x)
31(9(x)) → 91(3(x))
31(9(x)) → 31(x)

The TRS R consists of the following rules:

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

(4) Complex Obligation (AND)

(5) Obligation:

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

81(8(4(x))) → 91(x)
91(x) → 31(2(3(x)))
31(5(x)) → 81(9(7(x)))
31(5(x)) → 91(7(x))
91(x) → 21(3(x))
21(6(x)) → 31(x)
31(5(x)) → 71(x)
71(1(x)) → 91(x)
91(x) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(8(x)) → 71(x)
31(9(x)) → 91(3(x))
91(x) → 21(x)
31(9(x)) → 31(x)

The TRS R consists of the following rules:

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

81(8(4(x))) → 91(x)
91(x) → 31(2(3(x)))
31(5(x)) → 81(9(7(x)))
31(5(x)) → 91(7(x))
91(x) → 21(3(x))
21(6(x)) → 31(x)
31(5(x)) → 71(x)
71(1(x)) → 91(x)
91(x) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(8(x)) → 71(x)
31(9(x)) → 91(3(x))
91(x) → 21(x)
31(9(x)) → 31(x)

The TRS R consists of the following rules:

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


81(8(4(x))) → 91(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( 21(x1) ) = 0

POL( 81(x1) ) = max{0, 2x1 - 2}

POL( 91(x1) ) = max{0, -2}

POL( 2(x1) ) = max{0, -2}

POL( 31(x1) ) = max{0, -2}

POL( 4(x1) ) = max{0, -2}

POL( 3(x1) ) = 2x1

POL( 1(x1) ) = 2

POL( 5(x1) ) = 1

POL( 8(x1) ) = 2

POL( 9(x1) ) = 1

POL( 7(x1) ) = 0

POL( 0(x1) ) = max{0, -2}

POL( 6(x1) ) = 2

POL( 71(x1) ) = 0


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

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

(9) Obligation:

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

91(x) → 31(2(3(x)))
31(5(x)) → 81(9(7(x)))
31(5(x)) → 91(7(x))
91(x) → 21(3(x))
21(6(x)) → 31(x)
31(5(x)) → 71(x)
71(1(x)) → 91(x)
91(x) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(8(x)) → 71(x)
31(9(x)) → 91(3(x))
91(x) → 21(x)
31(9(x)) → 31(x)

The TRS R consists of the following rules:

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

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

(10) DependencyGraphProof (EQUIVALENT transformation)

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

(11) Obligation:

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

31(5(x)) → 91(7(x))
91(x) → 31(2(3(x)))
31(5(x)) → 71(x)
71(1(x)) → 91(x)
91(x) → 21(3(x))
21(6(x)) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(8(x)) → 71(x)
31(9(x)) → 91(3(x))
91(x) → 31(x)
31(9(x)) → 31(x)
91(x) → 21(x)

The TRS R consists of the following rules:

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

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

(12) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


71(1(x)) → 91(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(31(x1)) = 0A +
[0A,-I,0A]
·x1

POL(5(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A0A\
|0A1A-I|
\-I0A1A/
·x1

POL(91(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

POL(3(x1)) =
/0A\
|0A|
\-I/
 +
/-I0A-I\
|-I0A-I|
\-I0A-I/
·x1

POL(71(x1)) = 0A +
[0A,0A,1A]
·x1

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

POL(21(x1)) = 0A +
[0A,0A,0A]
·x1

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

POL(8(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A1A\
|-I-I0A|
\-I-I0A/
·x1

POL(9(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|0A0A-I|
\-I-I0A/
·x1

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

POL(4(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A-I\
|-I-I-I|
\0A0A0A/
·x1

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

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

(13) Obligation:

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

31(5(x)) → 91(7(x))
91(x) → 31(2(3(x)))
31(5(x)) → 71(x)
91(x) → 21(3(x))
21(6(x)) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(8(x)) → 71(x)
31(9(x)) → 91(3(x))
91(x) → 31(x)
31(9(x)) → 31(x)
91(x) → 21(x)

The TRS R consists of the following rules:

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

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(15) Obligation:

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

91(x) → 31(2(3(x)))
31(5(x)) → 91(7(x))
91(x) → 21(3(x))
21(6(x)) → 31(x)
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(9(x)) → 91(3(x))
91(x) → 31(x)
31(9(x)) → 31(x)
91(x) → 21(x)

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


21(6(x)) → 31(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,1A]
·x1

POL(31(x1)) = -I +
[0A,0A,0A]
·x1

POL(2(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A1A\
|-I0A0A|
\-I-I0A/
·x1

POL(3(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A1A\
|-I0A0A|
\-I-I0A/
·x1

POL(5(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A-I\
|-I-I-I|
\1A0A1A/
·x1

POL(7(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A1A\
|0A0A-I|
\0A-I-I/
·x1

POL(21(x1)) = 0A +
[-I,0A,1A]
·x1

POL(6(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|-I0A0A|
\0A0A0A/
·x1

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

POL(9(x1)) =
/0A\
|-I|
\1A/
 +
/0A0A1A\
|0A0A0A|
\0A0A1A/
·x1

POL(1(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A1A\
|-I-I-I|
\-I0A0A/
·x1

POL(4(x1)) =
/-I\
|0A|
\-I/
 +
/0A0A0A\
|-I0A0A|
\-I0A0A/
·x1

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

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

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

(17) Obligation:

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

91(x) → 31(2(3(x)))
31(5(x)) → 91(7(x))
91(x) → 21(3(x))
31(8(x)) → 31(2(7(x)))
31(8(x)) → 21(7(x))
31(9(x)) → 91(3(x))
91(x) → 31(x)
31(9(x)) → 31(x)
91(x) → 21(x)

The TRS R consists of the following rules:

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

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(19) Obligation:

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

31(5(x)) → 91(7(x))
91(x) → 31(2(3(x)))
31(8(x)) → 31(2(7(x)))
31(9(x)) → 91(3(x))
91(x) → 31(x)
31(9(x)) → 31(x)

The TRS R consists of the following rules:

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

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

(20) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


31(5(x)) → 91(7(x))
31(9(x)) → 91(3(x))
31(9(x)) → 31(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(3(x1)) = x1   
POL(31(x1)) = x1   
POL(4(x1)) = 0   
POL(5(x1)) = 1   
POL(6(x1)) = 0   
POL(7(x1)) = 0   
POL(8(x1)) = 0   
POL(9(x1)) = 1 + x1   
POL(91(x1)) = x1   

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

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

(21) Obligation:

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

91(x) → 31(2(3(x)))
31(8(x)) → 31(2(7(x)))
91(x) → 31(x)

The TRS R consists of the following rules:

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

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

(22) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(23) Obligation:

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

31(8(x)) → 31(2(7(x)))

The TRS R consists of the following rules:

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

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

(24) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


31(8(x)) → 31(2(7(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( 2(x1) ) = max{0, -2}

POL( 31(x1) ) = x1

POL( 4(x1) ) = max{0, -2}

POL( 8(x1) ) = 2

POL( 9(x1) ) = max{0, 2x1 - 2}

POL( 7(x1) ) = 2

POL( 1(x1) ) = max{0, -2}

POL( 6(x1) ) = x1 + 2

POL( 5(x1) ) = max{0, x1 - 2}

POL( 0(x1) ) = max{0, -2}

POL( 3(x1) ) = max{0, x1 - 2}


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

2(6(x)) → 4(3(x))

(25) Obligation:

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

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

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) YES

(28) Obligation:

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

51(9(x)) → 51(x)

The TRS R consists of the following rules:

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

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

(29) 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.

(30) Obligation:

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

51(9(x)) → 51(x)

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

(31) 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:

  • 51(9(x)) → 51(x)
    The graph contains the following edges 1 > 1

(32) YES