YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/09.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 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 51 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 AND
            ↳14 QDP
                ↳15 QDPOrderProof (⇔, 6 ms)
                ↳16 QDP
                ↳17 DependencyGraphProof (⇔, 0 ms)
                ↳18 TRUE
            ↳19 QDP
                ↳20 QDPOrderProof (⇔, 20 ms)
                ↳21 QDP
                ↳22 PisEmptyProof (⇔, 0 ms)
                ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPOrderProof (⇔, 9 ms)
        ↳28 QDP
        ↳29 MRRProof (⇔, 14 ms)
        ↳30 QDP
        ↳31 PisEmptyProof (⇔, 0 ms)
        ↳32 YES

(0) Obligation:

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

a(s(x)) → s(a(x))
b(a(b(s(x)))) → a(b(s(a(x))))
b(a(b(b(x)))) → c(s(x))
c(s(x)) → a(b(a(b(x))))
a(b(a(a(x)))) → b(a(b(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:

s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → s(c(x))
s(c(x)) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(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:

S(a(x)) → A(s(x))
S(a(x)) → S(x)
S(b(a(b(x)))) → A(s(b(a(x))))
S(b(a(b(x)))) → S(b(a(x)))
S(b(a(b(x)))) → B(a(x))
S(b(a(b(x)))) → A(x)
B(b(a(b(x)))) → S(c(x))
S(c(x)) → B(a(b(a(x))))
S(c(x)) → A(b(a(x)))
S(c(x)) → B(a(x))
S(c(x)) → A(x)
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)

The TRS R consists of the following rules:

s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → s(c(x))
s(c(x)) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(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 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

S(c(x)) → B(a(b(a(x))))
B(b(a(b(x)))) → S(c(x))
S(c(x)) → A(b(a(x)))
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
S(c(x)) → B(a(x))
S(c(x)) → A(x)

The TRS R consists of the following rules:

s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → s(c(x))
s(c(x)) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(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:

S(c(x)) → B(a(b(a(x))))
B(b(a(b(x)))) → S(c(x))
S(c(x)) → A(b(a(x)))
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
S(c(x)) → B(a(x))
S(c(x)) → A(x)

The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(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:

S(c(x)) → A(b(a(x)))
A(a(b(a(x)))) → A(b(x))
A(a(b(a(x)))) → B(x)
S(c(x)) → B(a(x))
S(c(x)) → A(x)


Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(B(x1)) = 1 + x1   
POL(S(x1)) = 2 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 2 + x1   
POL(s(x1)) = 2 + x1   

(11) Obligation:

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

S(c(x)) → B(a(b(a(x))))
B(b(a(b(x)))) → S(c(x))
A(a(b(a(x)))) → A(b(a(b(x))))
A(a(b(a(x)))) → B(a(b(x)))

The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(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 2 SCCs with 1 less node.

(13) Complex Obligation (AND)

(14) Obligation:

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

B(b(a(b(x)))) → S(c(x))
S(c(x)) → B(a(b(a(x))))

The TRS R consists of the following rules:

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


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

POL(B(x1)) = x1   
POL(S(x1)) = x1   
POL(a(x1)) = 0   
POL(b(x1)) = 1   
POL(c(x1)) = 0   
POL(s(x1)) = 1 + x1   

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

a(a(b(a(x)))) → a(b(a(b(x))))

(16) Obligation:

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

S(c(x)) → B(a(b(a(x))))

The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

A(a(b(a(x)))) → A(b(a(b(x))))

The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))

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

(20) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


A(a(b(a(x)))) → A(b(a(b(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( A(x1) ) = 2x1

POL( b(x1) ) = max{0, -2}

POL( a(x1) ) = 1

POL( s(x1) ) = max{0, -2}

POL( c(x1) ) = 2x1 + 2


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

b(b(a(b(x)))) → s(c(x))
s(c(x)) → b(a(b(a(x))))

(21) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))

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

(22) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(23) YES

(24) Obligation:

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

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

The TRS R consists of the following rules:

s(a(x)) → a(s(x))
s(b(a(b(x)))) → a(s(b(a(x))))
b(b(a(b(x)))) → s(c(x))
s(c(x)) → b(a(b(a(x))))
a(a(b(a(x)))) → a(b(a(b(x))))

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

(25) 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.

(26) Obligation:

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

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

The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))

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

(27) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


S(a(x)) → S(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(S(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1   
POL(c(x1)) = 1   
POL(s(x1)) = x1   

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

b(b(a(b(x)))) → s(c(x))
s(c(x)) → b(a(b(a(x))))

(28) Obligation:

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

S(b(a(b(x)))) → S(b(a(x)))

The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))

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

(29) 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:

S(b(a(b(x)))) → S(b(a(x)))


Used ordering: Polynomial interpretation [POLO]:

POL(S(x1)) = 2·x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   
POL(s(x1)) = 3 + x1   

(30) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a(a(b(a(x)))) → a(b(a(b(x))))
s(c(x)) → b(a(b(a(x))))
b(b(a(b(x)))) → s(c(x))

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

(31) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(32) YES