Termination w.r.t. Q proof of /tmp/tmpWXys0c/thiemann03.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SHUFF(x, y) → HEAD(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
SHUFFLE(x) → SHUFF(x, nil)
IF(false, x, y, z) → TAIL(x)
IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))
IF(false, x, y, z) → REVERSE(tail(x))
SHUFF(x, y) → NULL(x)
SHUFF(x, y) → APP(y, add(head(x), nil))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

SHUFF(x, y) → HEAD(x)
REVERSE(add(n, x)) → REVERSE(x)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
APP(add(n, x), y) → APP(x, y)
SHUFFLE(x) → SHUFF(x, nil)
IF(false, x, y, z) → TAIL(x)
IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))
IF(false, x, y, z) → REVERSE(tail(x))
SHUFF(x, y) → NULL(x)
SHUFF(x, y) → APP(y, add(head(x), nil))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(add(n, x), y) → APP(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (2)x_1
POL(add(x₁, x₂)) = 1/4 + (7/2)x_2

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

REVERSE(add(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

REVERSE(add(n, x)) → REVERSE(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 1/4 + (7/2)x_2
POL(REVERSE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(false, x, y, z) → SHUFF(reverse(tail(x)), z)
SHUFF(x, y) → IF(null(x), x, y, app(y, add(head(x), nil)))

The TRS R consists of the following rules:

null(nil) → true
null(add(n, x)) → false
tail(add(n, x)) → x
tail(nil) → nil
head(add(n, x)) → n
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(x) → shuff(x, nil)
shuff(x, y) → if(null(x), x, y, app(y, add(head(x), nil)))
if(true, x, y, z) → y
if(false, x, y, z) → shuff(reverse(tail(x)), z)

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