YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_05_SRS/jambox5.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 20 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 17 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 TRUE
    ↳14 QDP
        ↳15 QDPOrderProof (⇔, 183 ms)
        ↳16 QDP
        ↳17 QDPOrderProof (⇔, 286 ms)
        ↳18 QDP
        ↳19 DependencyGraphProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

a(b(x)) → d(a(a(x)))
c(a(x)) → b(b(x))
b(a(d(x))) → c(d(d(b(x))))
d(x) → a(x)
a(c(a(b(x)))) → 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 2 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

D(x) → A(x)
A(b(x)) → D(a(a(x)))
A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

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

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

(8) 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.

(9) Obligation:

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

D(x) → A(x)
A(b(x)) → D(a(a(x)))
A(b(x)) → A(a(x))
A(b(x)) → A(x)

The TRS R consists of the following rules:

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

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

(10) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A(b(x)) → D(a(a(x)))
A(b(x)) → A(a(x))
A(b(x)) → A(x)

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 3·x1   
POL(D(x1)) = 3·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 2 + 3·x1   
POL(c(x1)) = 3 + 3·x1   
POL(d(x1)) = x1   

(11) Obligation:

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

D(x) → A(x)

The TRS R consists of the following rules:

d(x) → a(x)

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

(12) DependencyGraphProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


C(a(x)) → B(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,-I,0A]
·x1

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

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

POL(C(x1)) = 1A +
[-I,0A,0A]
·x1

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

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

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

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

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


C(a(x)) → B(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:

POL(B(x1)) = 0A +
[-1A,-1A,-1A]
·x1

POL(a(x1)) =
/0A\
|0A|
\-1A/
 +
/-I1A-1A\
|-I-1A-1A|
\-1A-1A-1A/
·x1

POL(d(x1)) =
/0A\
|1A|
\-1A/
 +
/-I1A-1A\
|-1A-1A-1A|
\-1A-1A-1A/
·x1

POL(C(x1)) = 1A +
[-1A,-1A,1A]
·x1

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

POL(c(x1)) =
/1A\
|-1A|
\2A/
 +
/-1A-1A-1A\
|-I-I-I|
\-1A-1A2A/
·x1

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

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

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) Obligation:

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

B(a(d(x))) → B(x)

The TRS R consists of the following rules:

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

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

(21) 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.

(22) Obligation:

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

B(a(d(x))) → B(x)

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

(23) 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:

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

(24) YES