YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z092.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 46 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 2 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 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔, 0 ms)
        ↳33 YES

(0) Obligation:

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

q0(0(x)) → 0'(q1(x))
q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q1(1(x))) → q2(0(1'(x)))
0'(q1(1(x))) → q2(0'(1'(x)))
1'(q1(1(x))) → q2(1'(1'(x)))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q2(0'(x)) → 0'(q0(x))
q0(1'(x)) → 1'(q3(x))
q3(1'(x)) → 1'(q3(x))
q3(b(x)) → b(q4(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(0'(x1)) = x1   
POL(1(x1)) = 3 + x1   
POL(1'(x1)) = x1   
POL(b(x1)) = x1   
POL(q0(x1)) = 2 + x1   
POL(q1(x1)) = x1   
POL(q2(x1)) = 3 + x1   
POL(q3(x1)) = 1 + x1   
POL(q4(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

q0(0(x)) → 0'(q1(x))
q2(0'(x)) → 0'(q0(x))
q0(1'(x)) → 1'(q3(x))
q3(b(x)) → b(q4(x))


(2) Obligation:

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

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q1(1(x))) → q2(0(1'(x)))
0'(q1(1(x))) → q2(0'(1'(x)))
1'(q1(1(x))) → q2(1'(1'(x)))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(x))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(0'(x1)) = x1   
POL(1(x1)) = x1   
POL(1'(x1)) = x1   
POL(q1(x1)) = 1 + x1   
POL(q2(x1)) = x1   
POL(q3(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

0(q1(1(x))) → q2(0(1'(x)))
0'(q1(1(x))) → q2(0'(1'(x)))
1'(q1(1(x))) → q2(1'(1'(x)))


(4) Obligation:

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

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(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:

Q1(0(x)) → 01(q1(x))
Q1(0(x)) → Q1(x)
Q1(1'(x)) → 1'1(q1(x))
Q1(1'(x)) → Q1(x)
01(q2(0(x))) → 01(0(x))
0'1(q2(0(x))) → 0'1(0(x))
1'1(q2(0(x))) → 1'1(0(x))
01(q2(1'(x))) → 01(1'(x))
0'1(q2(1'(x))) → 0'1(1'(x))
1'1(q2(1'(x))) → 1'1(1'(x))
Q3(1'(x)) → 1'1(q3(x))
Q3(1'(x)) → Q3(x)

The TRS R consists of the following rules:

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(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 5 SCCs with 3 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

1'1(q2(1'(x))) → 1'1(1'(x))
1'1(q2(0(x))) → 1'1(0(x))

The TRS R consists of the following rules:

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(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:

1'1(q2(1'(x))) → 1'1(1'(x))
1'1(q2(0(x))) → 1'1(0(x))

The TRS R consists of the following rules:

0(q2(0(x))) → q2(0(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
1'(q2(0(x))) → q2(1'(0(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))

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:

  • 1'1(q2(1'(x))) → 1'1(1'(x))
    The graph contains the following edges 1 > 1

  • 1'1(q2(0(x))) → 1'1(0(x))
    The graph contains the following edges 1 > 1

(13) YES

(14) Obligation:

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

Q3(1'(x)) → Q3(x)

The TRS R consists of the following rules:

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(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:

Q3(1'(x)) → Q3(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:

  • Q3(1'(x)) → Q3(x)
    The graph contains the following edges 1 > 1

(18) YES

(19) Obligation:

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

0'1(q2(1'(x))) → 0'1(1'(x))
0'1(q2(0(x))) → 0'1(0(x))

The TRS R consists of the following rules:

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(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:

0'1(q2(1'(x))) → 0'1(1'(x))
0'1(q2(0(x))) → 0'1(0(x))

The TRS R consists of the following rules:

0(q2(0(x))) → q2(0(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
1'(q2(0(x))) → q2(1'(0(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))

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

(22) 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:

  • 0'1(q2(1'(x))) → 0'1(1'(x))
    The graph contains the following edges 1 > 1

  • 0'1(q2(0(x))) → 0'1(0(x))
    The graph contains the following edges 1 > 1

(23) YES

(24) Obligation:

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

01(q2(1'(x))) → 01(1'(x))
01(q2(0(x))) → 01(0(x))

The TRS R consists of the following rules:

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(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:

01(q2(1'(x))) → 01(1'(x))
01(q2(0(x))) → 01(0(x))

The TRS R consists of the following rules:

0(q2(0(x))) → q2(0(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
1'(q2(0(x))) → q2(1'(0(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))

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

(27) 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:

  • 01(q2(1'(x))) → 01(1'(x))
    The graph contains the following edges 1 > 1

  • 01(q2(0(x))) → 01(0(x))
    The graph contains the following edges 1 > 1

(28) YES

(29) Obligation:

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

Q1(1'(x)) → Q1(x)
Q1(0(x)) → Q1(x)

The TRS R consists of the following rules:

q1(0(x)) → 0(q1(x))
q1(1'(x)) → 1'(q1(x))
0(q2(0(x))) → q2(0(0(x)))
0'(q2(0(x))) → q2(0'(0(x)))
1'(q2(0(x))) → q2(1'(0(x)))
0(q2(1'(x))) → q2(0(1'(x)))
0'(q2(1'(x))) → q2(0'(1'(x)))
1'(q2(1'(x))) → q2(1'(1'(x)))
q3(1'(x)) → 1'(q3(x))

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

(30) 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.

(31) Obligation:

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

Q1(1'(x)) → Q1(x)
Q1(0(x)) → Q1(x)

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

(32) 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:

  • Q1(1'(x)) → Q1(x)
    The graph contains the following edges 1 > 1

  • Q1(0(x)) → Q1(x)
    The graph contains the following edges 1 > 1

(33) YES