YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z102-shift.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 85 ms)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 19 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QReductionProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPSizeChangeProof (⇔, 0 ms)
        ↳15 YES
    ↳16 QDP
        ↳17 UsableRulesProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QReductionProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QReductionProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES

(0) Obligation:

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

B(x) → W(M(M(M(V(x)))))
M(x) → x
M(V(a(x))) → V(Xa(x))
M(V(b(x))) → V(Xb(x))
M(V(c(x))) → V(Xc(x))
Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
Xa(E(x)) → a(E(x))
Xb(E(x)) → b(E(x))
Xc(E(x)) → c(E(x))
W(V(x)) → R(L(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
L(a(a(x))) → D(b(b(b(x))))
L(b(b(x))) → D(c(c(c(x))))
L(c(c(c(c(x))))) → D(a(b(x)))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))
R(D(x)) → B(x)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(B(x1)) = 8 + x1   
POL(D(x1)) = 9 + x1   
POL(E(x1)) = x1   
POL(L(x1)) = x1   
POL(M(x1)) = 2 + x1   
POL(R(x1)) = x1   
POL(V(x1)) = x1   
POL(W(x1)) = 1 + x1   
POL(Xa(x1)) = 261 + x1   
POL(Xb(x1)) = 171 + x1   
POL(Xc(x1)) = 111 + x1   
POL(Ya(x1)) = 260 + x1   
POL(Yb(x1)) = 170 + x1   
POL(Yc(x1)) = 110 + x1   
POL(a(x1)) = 260 + x1   
POL(b(x1)) = 170 + x1   
POL(c(x1)) = 110 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

B(x) → W(M(M(M(V(x)))))
M(x) → x
M(V(a(x))) → V(Xa(x))
M(V(b(x))) → V(Xb(x))
M(V(c(x))) → V(Xc(x))
Xa(E(x)) → a(E(x))
Xb(E(x)) → b(E(x))
Xc(E(x)) → c(E(x))
W(V(x)) → R(L(x))
L(a(a(x))) → D(b(b(b(x))))
L(b(b(x))) → D(c(c(c(x))))
L(c(c(c(c(x))))) → D(a(b(x)))
R(D(x)) → B(x)


(2) Obligation:

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

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

(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:

XA(a(x)) → XA(x)
XA(b(x)) → XA(x)
XA(c(x)) → XA(x)
XB(a(x)) → XB(x)
XB(b(x)) → XB(x)
XB(c(x)) → XB(x)
XC(a(x)) → XC(x)
XC(b(x)) → XC(x)
XC(c(x)) → XC(x)
L1(a(x)) → YA(L(x))
L1(a(x)) → L1(x)
L1(b(x)) → YB(L(x))
L1(b(x)) → L1(x)
L1(c(x)) → YC(L(x))
L1(c(x)) → L1(x)

The TRS R consists of the following rules:

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
L1(c(x)) → L1(x)

The TRS R consists of the following rules:

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(10) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
L1(c(x)) → L1(x)

R is empty.
The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(12) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

(13) Obligation:

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

L1(b(x)) → L1(x)
L1(a(x)) → L1(x)
L1(c(x)) → L1(x)

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

(14) 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:

  • L1(b(x)) → L1(x)
    The graph contains the following edges 1 > 1

  • L1(a(x)) → L1(x)
    The graph contains the following edges 1 > 1

  • L1(c(x)) → L1(x)
    The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

XC(b(x)) → XC(x)
XC(a(x)) → XC(x)
XC(c(x)) → XC(x)

The TRS R consists of the following rules:

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(17) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

XC(b(x)) → XC(x)
XC(a(x)) → XC(x)
XC(c(x)) → XC(x)

R is empty.
The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(19) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

(20) Obligation:

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

XC(b(x)) → XC(x)
XC(a(x)) → XC(x)
XC(c(x)) → XC(x)

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

(21) 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:

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

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

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

(22) YES

(23) Obligation:

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

XB(b(x)) → XB(x)
XB(a(x)) → XB(x)
XB(c(x)) → XB(x)

The TRS R consists of the following rules:

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(24) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(25) Obligation:

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

XB(b(x)) → XB(x)
XB(a(x)) → XB(x)
XB(c(x)) → XB(x)

R is empty.
The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(26) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

(27) Obligation:

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

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

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

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

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

(29) YES

(30) Obligation:

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

XA(b(x)) → XA(x)
XA(a(x)) → XA(x)
XA(c(x)) → XA(x)

The TRS R consists of the following rules:

Xa(a(x)) → a(Xa(x))
Xa(b(x)) → b(Xa(x))
Xa(c(x)) → c(Xa(x))
Xb(a(x)) → a(Xb(x))
Xb(b(x)) → b(Xb(x))
Xb(c(x)) → c(Xb(x))
Xc(a(x)) → a(Xc(x))
Xc(b(x)) → b(Xc(x))
Xc(c(x)) → c(Xc(x))
L(a(x)) → Ya(L(x))
L(b(x)) → Yb(L(x))
L(c(x)) → Yc(L(x))
Ya(D(x)) → D(a(x))
Yb(D(x)) → D(b(x))
Yc(D(x)) → D(c(x))

The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(31) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(32) Obligation:

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

XA(b(x)) → XA(x)
XA(a(x)) → XA(x)
XA(c(x)) → XA(x)

R is empty.
The set Q consists of the following terms:

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

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

(33) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Xa(a(x0))
Xa(b(x0))
Xa(c(x0))
Xb(a(x0))
Xb(b(x0))
Xb(c(x0))
Xc(a(x0))
Xc(b(x0))
Xc(c(x0))
L(a(x0))
L(b(x0))
L(c(x0))
Ya(D(x0))
Yb(D(x0))
Yc(D(x0))

(34) Obligation:

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

XA(b(x)) → XA(x)
XA(a(x)) → XA(x)
XA(c(x)) → XA(x)

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

(35) 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:

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

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

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

(36) YES