YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z091.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 1 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 QDPOrderProof (⇔, 56 ms)
        ↳32 QDP
        ↳33 QDPOrderProof (⇔, 19 ms)
        ↳34 QDP
        ↳35 PisEmptyProof (⇔, 0 ms)
        ↳36 YES

(0) Obligation:

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

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

R0(0(x)) → 01(r0(x))
R0(0(x)) → R0(x)
R0(1(x)) → 11(r0(x))
R0(1(x)) → R0(x)
R0(m(x)) → M(r0(x))
R0(m(x)) → R0(x)
R1(0(x)) → 01(r1(x))
R1(0(x)) → R1(x)
R1(1(x)) → 11(r1(x))
R1(1(x)) → R1(x)
R1(m(x)) → M(r1(x))
R1(m(x)) → R1(x)
R0(b(x)) → 01(b(x))
R1(b(x)) → 11(b(x))
01(qr(x)) → 01(x)
11(qr(x)) → 11(x)
M(qr(x)) → M(x)
01(ql(x)) → 01(x)
11(ql(x)) → 11(x)
B(ql(0(x))) → 01(b(r0(x)))
B(ql(0(x))) → B(r0(x))
B(ql(0(x))) → R0(x)
B(ql(1(x))) → 11(b(r1(x)))
B(ql(1(x))) → B(r1(x))
B(ql(1(x))) → R1(x)

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 12 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

M(qr(x)) → M(x)

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(6) 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.

(7) Obligation:

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

M(qr(x)) → M(x)

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

(8) 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:

  • M(qr(x)) → M(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

11(ql(x)) → 11(x)
11(qr(x)) → 11(x)

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(11) 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.

(12) Obligation:

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

11(ql(x)) → 11(x)
11(qr(x)) → 11(x)

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

(13) 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:

  • 11(ql(x)) → 11(x)
    The graph contains the following edges 1 > 1

  • 11(qr(x)) → 11(x)
    The graph contains the following edges 1 > 1

(14) YES

(15) Obligation:

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

01(ql(x)) → 01(x)
01(qr(x)) → 01(x)

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(16) 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.

(17) Obligation:

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

01(ql(x)) → 01(x)
01(qr(x)) → 01(x)

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

(18) 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(ql(x)) → 01(x)
    The graph contains the following edges 1 > 1

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

(19) YES

(20) Obligation:

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

R1(1(x)) → R1(x)
R1(0(x)) → R1(x)
R1(m(x)) → R1(x)

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(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:

R1(1(x)) → R1(x)
R1(0(x)) → R1(x)
R1(m(x)) → R1(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:

  • R1(1(x)) → R1(x)
    The graph contains the following edges 1 > 1

  • R1(0(x)) → R1(x)
    The graph contains the following edges 1 > 1

  • R1(m(x)) → R1(x)
    The graph contains the following edges 1 > 1

(24) YES

(25) Obligation:

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

R0(1(x)) → R0(x)
R0(0(x)) → R0(x)
R0(m(x)) → R0(x)

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(26) 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.

(27) Obligation:

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

R0(1(x)) → R0(x)
R0(0(x)) → R0(x)
R0(m(x)) → R0(x)

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

(28) 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:

  • R0(1(x)) → R0(x)
    The graph contains the following edges 1 > 1

  • R0(0(x)) → R0(x)
    The graph contains the following edges 1 > 1

  • R0(m(x)) → R0(x)
    The graph contains the following edges 1 > 1

(29) YES

(30) Obligation:

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

B(ql(1(x))) → B(r1(x))
B(ql(0(x))) → B(r0(x))

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(31) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(ql(1(x))) → B(r1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = x1   
POL(1(x1)) = 1 + x1   
POL(B(x1)) = x1   
POL(b(x1)) = 0   
POL(m(x1)) = 1   
POL(ql(x1)) = x1   
POL(qr(x1)) = 0   
POL(r0(x1)) = x1   
POL(r1(x1)) = x1   

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

r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r1(b(x)) → qr(1(b(x)))
r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r0(b(x)) → qr(0(b(x)))
1(qr(x)) → qr(1(x))
1(ql(x)) → ql(1(x))
0(qr(x)) → qr(0(x))
0(ql(x)) → ql(0(x))
m(qr(x)) → ql(m(x))

(32) Obligation:

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

B(ql(0(x))) → B(r0(x))

The TRS R consists of the following rules:

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(33) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(ql(0(x))) → B(r0(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(1(x1)) = 0   
POL(B(x1)) = x1   
POL(b(x1)) = 0   
POL(m(x1)) = 1   
POL(ql(x1)) = x1   
POL(qr(x1)) = 0   
POL(r0(x1)) = x1   
POL(r1(x1)) = 0   

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

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r0(b(x)) → qr(0(b(x)))
1(qr(x)) → qr(1(x))
1(ql(x)) → ql(1(x))
0(qr(x)) → qr(0(x))
0(ql(x)) → ql(0(x))
m(qr(x)) → ql(m(x))

(34) Obligation:

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

r0(0(x)) → 0(r0(x))
r0(1(x)) → 1(r0(x))
r0(m(x)) → m(r0(x))
r1(0(x)) → 0(r1(x))
r1(1(x)) → 1(r1(x))
r1(m(x)) → m(r1(x))
r0(b(x)) → qr(0(b(x)))
r1(b(x)) → qr(1(b(x)))
0(qr(x)) → qr(0(x))
1(qr(x)) → qr(1(x))
m(qr(x)) → ql(m(x))
0(ql(x)) → ql(0(x))
1(ql(x)) → ql(1(x))
b(ql(0(x))) → 0(b(r0(x)))
b(ql(1(x))) → 1(b(r1(x)))

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) YES