Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE
  ↳22 QDP

(0) Obligation:

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

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

FROM(X) → FROM(s(X))
SEL(s(N), cons(X, XS)) → SEL(N, 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)) → ZWQUOT(XS, YS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(8) 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)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
s1: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

ZWQUOT(cons(X, XS), cons(Y, YS)) → ZWQUOT(XS, YS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(13) 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)) → ZWQUOT(XS, YS)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWQUOT(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
cons2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

SEL(s(N), cons(X, XS)) → SEL(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(18) 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, XS)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
s1 > SEL2
cons2 > SEL2

Status:
cons2: [1,2]
s1: [1]
SEL2: [2,1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, 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), zWquot(XS, YS))

The set Q consists of the following terms:

from(x0)
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
zWquot(x0, nil)
zWquot(nil, x0)
zWquot(cons(x0, x1), cons(x2, x3))

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