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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(s(X), s(Y)) → MINUS(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
s1 > MINUS1

Status:
MINUS1: [1]
s1: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
0  =  0

Lexicographic path order with status [LPO].
Precedence:
QUOT2 > minus1 > 0
s1 > minus1 > 0

Status:
QUOT2: [1,2]
s1: [1]
minus1: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

Q DP problem:
P is empty.
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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWQUOT(x1, x2)  =  ZWQUOT(x1, x2)
cons(x1, x2)  =  cons(x2)
ACTIVATE(x1)  =  x1
n__zWquot(x1, x2)  =  n__zWquot(x1, x2)

Lexicographic path order with status [LPO].
Precedence:
nzWquot2 > ZWQUOT2

Status:
ZWQUOT2: [1,2]
cons1: [1]
nzWquot2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

Q DP problem:
P is empty.
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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
activate(x1)  =  activate(x1)
n__from(x1)  =  n__from(x1)
from(x1)  =  from(x1)
n__zWquot(x1, x2)  =  n__zWquot(x2)
zWquot(x1, x2)  =  zWquot(x2)
nil  =  nil
quot(x1, x2)  =  quot
0  =  0
minus(x1, x2)  =  minus

Lexicographic path order with status [LPO].
Precedence:
SEL1 > activate1
cons2 > activate1
nfrom1 > activate1
from1 > s1 > minus > activate1
nzWquot1 > zWquot1 > nil > activate1
quot > s1 > minus > activate1
0 > activate1

Status:
SEL1: [1]
s1: [1]
cons2: [1,2]
activate1: [1]
nfrom1: [1]
from1: [1]
nzWquot1: [1]
zWquot1: [1]
nil: []
quot: []
0: []
minus: []

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

Q DP problem:
P is empty.
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.

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE