YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_05_SRS/aprove2.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 26 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPOrderProof (⇔, 142 ms)
        ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES

(0) Obligation:

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

g(h(x)) → g(f(s(x)))
f(s(s(s(x)))) → h(f(s(h(x))))
f(h(x)) → h(f(s(h(x))))
h(x) → x
f(f(s(s(x)))) → s(s(s(f(f(x)))))
b(a(x)) → a(b(x))
a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(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:

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))
a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(x)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = 3 + x1   
POL(b(x1)) = 1 + x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(h(x1)) = x1   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a(a(a(x))) → b(a(a(b(x))))
b(b(b(b(x)))) → a(x)


(4) Obligation:

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

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(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:

H(g(x)) → S(f(g(x)))
S(s(s(f(x)))) → H(s(f(h(x))))
S(s(s(f(x)))) → S(f(h(x)))
S(s(s(f(x)))) → H(x)
H(f(x)) → H(s(f(h(x))))
H(f(x)) → S(f(h(x)))
H(f(x)) → H(x)
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(s(x))
S(s(f(f(x)))) → S(x)
A(b(x)) → A(x)

The TRS R consists of the following rules:

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(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 3 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

A(b(x)) → A(x)

The TRS R consists of the following rules:

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))

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

(10) 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.

(11) Obligation:

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

A(b(x)) → A(x)

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

(12) 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:

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

(13) YES

(14) Obligation:

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

H(f(x)) → H(x)

The TRS R consists of the following rules:

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))

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

(15) 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.

(16) Obligation:

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

H(f(x)) → H(x)

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

(17) 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:

  • H(f(x)) → H(x)
    The graph contains the following edges 1 > 1

(18) YES

(19) Obligation:

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

S(s(f(f(x)))) → S(s(x))
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)

The TRS R consists of the following rules:

h(g(x)) → s(f(g(x)))
s(s(s(f(x)))) → h(s(f(h(x))))
h(f(x)) → h(s(f(h(x))))
h(x) → x
s(s(f(f(x)))) → f(f(s(s(s(x)))))
a(b(x)) → b(a(x))

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

(20) 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.

(21) Obligation:

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

S(s(f(f(x)))) → S(s(x))
S(s(f(f(x)))) → S(s(s(x)))
S(s(f(f(x)))) → S(x)

The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(g(x)) → s(f(g(x)))
h(f(x)) → h(s(f(h(x))))
h(x) → x

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

(22) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(S(x1)) = 0A +
[-I,0A,0A]
·x1

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

POL(f(x1)) =
/0A\
|-I|
\1A/
 +
/0A0A0A\
|-I0A0A|
\-I1A-I/
·x1

POL(h(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A0A\
|-I0A-I|
\-I0A0A/
·x1

POL(g(x1)) =
/-I\
|-I|
\-I/
 +
/1A1A0A\
|0A0A-I|
\0A0A-I/
·x1

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

s(s(s(f(x)))) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(g(x)) → s(f(g(x)))
h(f(x)) → h(s(f(h(x))))
h(x) → x

(23) Obligation:

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

S(s(f(f(x)))) → S(x)

The TRS R consists of the following rules:

s(s(s(f(x)))) → h(s(f(h(x))))
s(s(f(f(x)))) → f(f(s(s(s(x)))))
h(g(x)) → s(f(g(x)))
h(f(x)) → h(s(f(h(x))))
h(x) → x

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

(24) 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.

(25) Obligation:

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

S(s(f(f(x)))) → S(x)

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

(26) 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:

  • S(s(f(f(x)))) → S(x)
    The graph contains the following edges 1 > 1

(27) YES