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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
    ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔)
        ↳16 TRUE

(0) Obligation:

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

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(X)
activate(n__zWquot(X1, X2)) → zWquot(X1, X2)
activate(X) → 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:

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
MINUS(s(X), s(Y)) → MINUS(X, Y)
QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) → MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) → QUOT(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(X1, X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(X)
activate(n__zWquot(X1, X2)) → zWquot(X1, X2)
activate(X) → 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 4 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(X)
activate(n__zWquot(X1, X2)) → zWquot(X1, X2)
activate(X) → X

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • MINUS(s(X), s(Y)) → MINUS(X, Y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) Obligation:

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

QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(X)
activate(n__zWquot(X1, X2)) → zWquot(X1, X2)
activate(X) → X

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:


POL(0) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

From the DPs we obtained the following set of size-change graphs:

  • QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y)) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05].


minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)

(10) TRUE

(11) Obligation:

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

ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS)
ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(X1, X2)
ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(X)
activate(n__zWquot(X1, X2)) → zWquot(X1, X2)
activate(X) → X

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
cons(x1, x2)  =  x2
n__zWquot(x1, x2)  =  n__zWquot(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • ACTIVATE(n__zWquot(X1, X2)) → ZWQUOT(X1, X2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 1 > 2

  • ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(XS) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 >= 1

  • ZWQUOT(cons(X, XS), cons(Y, YS)) → ACTIVATE(YS) (allowed arguments on rhs = {1})
    The graph contains the following edges 2 >= 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(13) TRUE

(14) Obligation:

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

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
minus(X, 0) → 0
minus(s(X), s(Y)) → minus(X, Y)
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), n__zWquot(activate(XS), activate(YS)))
from(X) → n__from(X)
zWquot(X1, X2) → n__zWquot(X1, X2)
activate(n__from(X)) → from(X)
activate(n__zWquot(X1, X2)) → zWquot(X1, X2)
activate(X) → X

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)
cons(x1, x2)  =  cons

From the DPs we obtained the following set of size-change graphs:

  • SEL(s(N), cons(X, XS)) → SEL(N, activate(XS)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(16) TRUE