Termination proof of /tmp/tmpMJV_PZ/Ex4_7_37

Termination of the following Term Rewriting System could be proven:

Context-sensitive 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 replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
sel: {1, 2}
0: empty set
minus: {1, 2}
quot: {1, 2}
zWquot: {1, 2}
nil: empty set

↳ CSR

  ↳ CSDependencyPairsProof

Context-sensitive 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 replacement map contains the following entries:

from: {1}
cons: {1}
s: {1}
sel: {1, 2}
0: empty set
minus: {1, 2}
quot: {1, 2}
zWquot: {1, 2}
nil: empty set

Using Improved CS-DPs we result in the following initial Q-CSDP problem.

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, sel, minus, quot, zWquot, SEL, MINUS, QUOT, ZWQUOT, FROM} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The ordinary context-sensitive dependency pairs DP_o are:

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)

The collapsing dependency pairs are DP_c:

SEL(s(N), cons(X, XS)) → XS

The hidden terms of R are:

from(s(X))
zWquot(XS, YS)

Every hiding context is built from:

s on positions {1}
from on positions {1}
zWquot on positions {1, 2}

Hence, the new unhiding pairs DP_u are :

SEL(s(N), cons(X, XS)) → U(XS)
U(s(x_0)) → U(x_0)
U(from(x_0)) → U(x_0)
U(zWquot(x_0, x_1)) → U(x_0)
U(zWquot(x_0, x_1)) → U(x_1)
U(from(s(X))) → FROM(s(X))
U(zWquot(XS, 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))

Q is empty.

The approximation of the Context-Sensitive Dependency Graph contains 4 SCCs with 5 less nodes.

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

            ↳ QCSDPSubtermProof

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, sel, minus, quot, zWquot, MINUS} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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))

Q is empty.

We use the subterm processor [20].

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.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = x1

Subterm Order

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

            ↳ QCSDPSubtermProof

              ↳ QCSDP

                ↳ PIsEmptyProof

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, sel, minus, quot, zWquot} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

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.

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

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

          ↳ QCSDP

            ↳ QCSDPReductionPairProof

          ↳ QCSDP

          ↳ QCSDP

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, sel, minus, quot, zWquot, QUOT} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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, 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.

Using the order
Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(QUOT(x₁, x₂)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(from(x₁)) = 1 + x₁
POL(minus(x₁, x₂)) = 0
POL(nil) = 1
POL(quot(x₁, x₂)) = 1
POL(s(x₁)) = 1
POL(zWquot(x₁, x₂)) = 1 + x₂

the following usable rules

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)))
from(X) → cons(X, from(s(X)))
zWquot(XS, nil) → nil
zWquot(nil, XS) → nil
zWquot(cons(X, XS), cons(Y, YS)) → cons(quot(X, Y), zWquot(XS, YS))

could all be oriented weakly.
Furthermore, the pairs

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

could be oriented strictly and thus removed.
All pairs have been removed.

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

          ↳ QCSDP

            ↳ QCSDPReductionPairProof

              ↳ QCSDP

                ↳ PIsEmptyProof

          ↳ QCSDP

          ↳ QCSDP

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.

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

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

            ↳ QCSDPSubtermProof

          ↳ QCSDP

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, sel, minus, quot, zWquot} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.
The symbols in {U} are not replacing on any position.

The TRS P consists of the following rules:

U(s(x_0)) → U(x_0)
U(from(x_0)) → U(x_0)
U(zWquot(x_0, x_1)) → U(x_0)
U(zWquot(x_0, x_1)) → U(x_1)

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.

We use the subterm processor [20].

The following pairs can be oriented strictly and are deleted.

U(s(x_0)) → U(x_0)
U(from(x_0)) → U(x_0)
U(zWquot(x_0, x_1)) → U(x_0)
U(zWquot(x_0, x_1)) → U(x_1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
U(x1) = x1

Subterm Order

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

            ↳ QCSDPSubtermProof

              ↳ QCSDP

                ↳ PIsEmptyProof

          ↳ QCSDP

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.

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

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

            ↳ QCSDPSubtermProof

Q-restricted context-sensitive dependency pair problem:
The symbols in {from, s, sel, minus, quot, zWquot, SEL} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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))

Q is empty.

We use the subterm processor [20].

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.
none
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2) = x1

Subterm Order

↳ CSR

  ↳ CSDependencyPairsProof

    ↳ QCSDP

      ↳ QCSDependencyGraphProof

        ↳ AND

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

          ↳ QCSDP

            ↳ QCSDPSubtermProof

              ↳ QCSDP

                ↳ PIsEmptyProof

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.

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