YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_06_SRS/secr4.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 21 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 270 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 73 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 128 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 258 ms)
        ↳15 QDP
        ↳16 DependencyGraphProof (⇔, 0 ms)
        ↳17 TRUE
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

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

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

A(a(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
A(a(x)) → B(x)
A(x) → C(d(x))
B(b(x)) → C(c(c(x)))
B(b(x)) → C(c(x))
B(b(x)) → C(x)
E(d(x)) → A(b(c(d(e(x)))))
E(d(x)) → B(c(d(e(x))))
E(d(x)) → C(d(e(x)))
E(d(x)) → E(x)
E(c(x)) → B(a(a(e(x))))
E(c(x)) → A(a(e(x)))
E(c(x)) → A(e(x))
E(c(x)) → E(x)
C(d(d(x))) → A(x)

The TRS R consists of the following rules:

a(a(x)) → b(b(b(x)))
a(x) → c(d(x))
b(b(x)) → c(c(c(x)))
c(c(x)) → d(d(d(x)))
e(d(x)) → a(b(c(d(e(x)))))
b(x) → d(d(x))
e(c(x)) → b(a(a(e(x))))
c(d(d(x))) → a(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 6 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

POL(B(x1)) = 0A +
[-I,0A,0A]
·x1

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

POL(C(x1)) = 0A +
[-I,-I,0A]
·x1

POL(c(x1)) =
/0A\
|-I|
\0A/
 +
/-I0A1A\
|0A0A0A|
\-I0A-I/
·x1

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

POL(A(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

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

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(b(x)) → C(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,0A,0A]
·x1

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

POL(C(x1)) = -I +
[0A,0A,-I]
·x1

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

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

POL(A(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

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

(11) Obligation:

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

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

The TRS R consists of the following rules:

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


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

POL(B(x1)) = -I +
[0A,0A,0A]
·x1

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

POL(C(x1)) = 0A +
[-I,0A,-I]
·x1

POL(c(x1)) =
/1A\
|-I|
\-I/
 +
/0A1A0A\
|-I0A0A|
\0A0A0A/
·x1

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

POL(A(x1)) = 0A +
[0A,0A,0A]
·x1

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

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

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

(13) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


C(d(d(x))) → A(x)
A(a(x)) → B(b(b(x)))
A(x) → C(d(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(A(x1)) = [15/4] + [2]x1   
POL(B(x1)) = [3/2] + [2]x1   
POL(C(x1)) = [2]x1   
POL(a(x1)) = [7/2] + x1   
POL(b(x1)) = [9/4] + x1   
POL(c(x1)) = [3/2] + x1   
POL(d(x1)) = [1] + x1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(15) Obligation:

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

B(b(x)) → C(c(c(x)))

The TRS R consists of the following rules:

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

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

(16) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(17) TRUE

(18) Obligation:

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

E(c(x)) → E(x)
E(d(x)) → E(x)

The TRS R consists of the following rules:

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

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

(19) 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.

(20) Obligation:

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

E(c(x)) → E(x)
E(d(x)) → E(x)

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

(21) 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:

  • E(c(x)) → E(x)
    The graph contains the following edges 1 > 1

  • E(d(x)) → E(x)
    The graph contains the following edges 1 > 1

(22) YES