YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_07_SRS/x05.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 UsableRulesProof (⇔, 0 ms)
↳8 QDP
↳9 MNOCProof (⇔, 0 ms)
↳10 QDP
↳11 QDPOrderProof (⇔, 102 ms)
↳12 QDP
↳13 QDPOrderProof (⇔, 72 ms)
↳14 QDP
↳15 DependencyGraphProof (⇔, 0 ms)
↳16 TRUE

(0) Obligation:

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

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

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

A(a(x)) → B(b(x))
A(a(x)) → B(x)
B(c(c(x))) → A(c(d(x)))
B(c(c(x))) → D(x)
A(x) → D(x)
D(c(x)) → B(x)

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

A(x) → D(x)
D(c(x)) → B(x)
B(c(c(x))) → A(c(d(x)))
B(c(c(x))) → D(x)

The TRS R consists of the following rules:

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

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

(7) 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.

(8) Obligation:

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

A(x) → D(x)
D(c(x)) → B(x)
B(c(c(x))) → A(c(d(x)))
B(c(c(x))) → D(x)

The TRS R consists of the following rules:

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

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

(9) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(10) Obligation:

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

A(x) → D(x)
D(c(x)) → B(x)
B(c(c(x))) → A(c(d(x)))
B(c(c(x))) → D(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

d(c(x0))
b(c(c(x0)))
a(x0)

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.


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

POL(A(x1)) = 0A +
[-I,0A,0A]
·x1

POL(D(x1)) = 0A +
[-I,0A,0A]
·x1

POL(c(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A1A\
|0A0A-I|
\-I0A0A/
·x1

POL(B(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

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

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

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

(12) Obligation:

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

A(x) → D(x)
D(c(x)) → B(x)
B(c(c(x))) → A(c(d(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

d(c(x0))
b(c(c(x0)))
a(x0)

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.


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

POL(A(x1)) = 1A +
[0A,0A,1A]
·x1

POL(D(x1)) = 0A +
[-I,-I,0A]
·x1

POL(c(x1)) =
/0A\
|1A|
\-I/
 +
/0A0A0A\
|0A0A1A|
\0A-I0A/
·x1

POL(B(x1)) = 0A +
[0A,-I,0A]
·x1

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

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

POL(a(x1)) =
/1A\
|1A|
\1A/
 +
/0A0A1A\
|-I0A1A|
\-I0A1A/
·x1

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

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

(14) Obligation:

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

D(c(x)) → B(x)
B(c(c(x))) → A(c(d(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

d(c(x0))
b(c(c(x0)))
a(x0)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE