YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z100.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 QDPOrderProof (⇔, 11 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳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 QDPOrderProof (⇔, 474 ms)
        ↳24 QDP
        ↳25 QDPOrderProof (⇔, 10 ms)
        ↳26 QDP
        ↳27 PisEmptyProof (⇔, 0 ms)
        ↳28 YES

(0) Obligation:

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

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → 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:

R1(a(x)) → A(a(a(r1(x))))
R1(a(x)) → A(a(r1(x)))
R1(a(x)) → A(r1(x))
R1(a(x)) → R1(x)
R2(a(x)) → A(a(a(r2(x))))
R2(a(x)) → A(a(r2(x)))
R2(a(x)) → A(r2(x))
R2(a(x)) → R2(x)
A(l1(x)) → A(a(a(x)))
A(l1(x)) → A(a(x))
A(l1(x)) → A(x)
A(a(l2(x))) → A(a(x))
A(a(l2(x))) → A(x)
R2(b(x)) → A(b(x))
B(l1(x)) → B(r2(x))
B(l1(x)) → R2(x)
B(l2(x)) → B(r1(x))
B(l2(x)) → R1(x)

The TRS R consists of the following rules:

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → 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 4 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A(l1(x)) → A(a(x))
A(l1(x)) → A(a(a(x)))
A(l1(x)) → A(x)
A(a(l2(x))) → A(a(x))
A(a(l2(x))) → A(x)

The TRS R consists of the following rules:

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → 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(l1(x)) → A(a(x))
A(l1(x)) → A(a(a(x)))
A(l1(x)) → A(x)
A(a(l2(x))) → A(a(x))
A(a(l2(x))) → A(x)

The TRS R consists of the following rules:

a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
a(a(x)) → 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(l1(x)) → A(a(x))
A(l1(x)) → A(a(a(x)))
A(l1(x)) → A(x)
A(a(l2(x))) → A(a(x))
A(a(l2(x))) → A(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(a(x1)) = x1   
POL(l1(x1)) = 1 + x1   
POL(l2(x1)) = 1 + x1   

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

a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
a(a(x)) → x

(9) Obligation:

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

a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
a(a(x)) → 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:

R2(a(x)) → R2(x)

The TRS R consists of the following rules:

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → 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:

R2(a(x)) → R2(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:

  • R2(a(x)) → R2(x)
    The graph contains the following edges 1 > 1

(16) YES

(17) Obligation:

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

R1(a(x)) → R1(x)

The TRS R consists of the following rules:

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → 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:

R1(a(x)) → R1(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:

  • R1(a(x)) → R1(x)
    The graph contains the following edges 1 > 1

(21) YES

(22) Obligation:

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

B(l2(x)) → B(r1(x))
B(l1(x)) → B(r2(x))

The TRS R consists of the following rules:

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → 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.


B(l1(x)) → B(r2(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x1)) = 0A +
[0A,-I,-I]
·x1

POL(l2(x1)) =
/1A\
|0A|
\0A/
 +
/0A1A0A\
|-I0A-I|
\-I0A-I/
·x1

POL(r1(x1)) =
/0A\
|0A|
\0A/
 +
/0A1A0A\
|1A0A0A|
\0A0A1A/
·x1

POL(l1(x1)) =
/1A\
|1A|
\-I/
 +
/0A1A0A\
|1A0A0A|
\0A0A0A/
·x1

POL(r2(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|0A-I-I|
\0A-I0A/
·x1

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

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

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

r1(a(x)) → a(a(a(r1(x))))
r1(b(x)) → l1(b(x))
r2(a(x)) → a(a(a(r2(x))))
r2(b(x)) → l2(a(b(x)))
b(l2(x)) → b(r1(x))
b(l1(x)) → b(r2(x))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
a(a(x)) → x

(24) Obligation:

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

B(l2(x)) → B(r1(x))

The TRS R consists of the following rules:

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → x

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

(25) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(l2(x)) → B(r1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(B(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(l1(x1)) = 0   
POL(l2(x1)) = 1   
POL(r1(x1)) = 0   
POL(r2(x1)) = 1 + x1   

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

r1(a(x)) → a(a(a(r1(x))))
r1(b(x)) → l1(b(x))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
a(a(x)) → x

(26) Obligation:

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

r1(a(x)) → a(a(a(r1(x))))
r2(a(x)) → a(a(a(r2(x))))
a(l1(x)) → l1(a(a(a(x))))
a(a(l2(x))) → l2(a(a(x)))
r1(b(x)) → l1(b(x))
r2(b(x)) → l2(a(b(x)))
b(l1(x)) → b(r2(x))
b(l2(x)) → b(r1(x))
a(a(x)) → x

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

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) YES