YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_05_SRS/torpa4.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 41 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 30 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 35 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 UsableRulesProof (⇔, 0 ms)
↳14 QDP
↳15 QDPSizeChangeProof (⇔, 0 ms)
↳16 YES

(0) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = 12 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = 18 + x1   
POL(d(x1)) = 7 + x1   
POL(f(x1)) = 9 + x1   
POL(g(x1)) = 30 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

d(a(x)) → c(x)
d(c(x)) → a(a(x))
g(x) → d(d(d(d(x))))


(4) Obligation:

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

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
g(x) → a(c(x))

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

A(c(b(a(x)))) → A(b(c(a(b(x)))))
A(c(b(a(x)))) → C(a(b(x)))
A(c(b(a(x)))) → A(b(x))
F(f(a(x))) → G(x)
G(b(x)) → G(x)
C(x) → F(f(x))
C(x) → F(x)
C(a(c(x))) → C(b(a(c(b(x)))))
C(a(c(x))) → A(c(b(x)))
C(a(c(x))) → C(b(x))
G(x) → A(c(x))
G(x) → C(x)

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
g(x) → a(c(x))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

A(c(b(a(x)))) → C(a(b(x)))
C(x) → F(f(x))
F(f(a(x))) → G(x)
G(b(x)) → G(x)
G(x) → A(c(x))
G(x) → C(x)
C(x) → F(x)
C(a(c(x))) → C(b(a(c(b(x)))))
C(a(c(x))) → A(c(b(x)))
C(a(c(x))) → C(b(x))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
g(x) → a(c(x))

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

(9) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(A(x1)) = x1   
POL(C(x1)) = x1   
POL(F(x1)) = x1   
POL(G(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(f(x1)) = x1   
POL(g(x1)) = 1 + x1   

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

f(f(a(x))) → g(x)
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
g(b(x)) → b(g(x))
g(x) → a(c(x))

(10) Obligation:

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

A(c(b(a(x)))) → C(a(b(x)))
C(x) → F(f(x))
F(f(a(x))) → G(x)
G(b(x)) → G(x)
C(x) → F(x)
C(a(c(x))) → C(b(a(c(b(x)))))

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
g(x) → a(c(x))

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(12) Obligation:

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

G(b(x)) → G(x)

The TRS R consists of the following rules:

a(c(b(a(x)))) → b(a(b(c(a(b(x))))))
f(f(a(x))) → g(x)
g(b(x)) → b(g(x))
c(x) → f(f(x))
c(a(c(x))) → c(b(a(c(b(x)))))
g(x) → a(c(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:

G(b(x)) → G(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:

  • G(b(x)) → G(x)
    The graph contains the following edges 1 > 1

(16) YES