Termination w.r.t. Q proof of /tmp/tmp5yuIei/Ex1SimplyTyped.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

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:

APP(app(add, app(s, x)), y) → APP(app(add, x), y)
APP(app(add, app(s, x)), y) → APP(add, x)
APP(app(add, app(s, x)), y) → APP(s, app(app(add, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

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:

APP(app(add, app(s, x)), y) → APP(app(add, x), y)
APP(app(add, app(s, x)), y) → APP(add, x)
APP(app(add, app(s, x)), y) → APP(s, app(app(add, x), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))

The TRS R consists of the following rules:

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

APP(app(add, app(s, x)), y) → APP(app(add, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

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(app(add, app(s, x)), y) → APP(app(add, x), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)

The remaining pairs can at least be oriented weakly.

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (1/2)x_1 + (4)x_2
POL(add) = 0
POL(cons) = 0
POL(map) = 7/2
POL(app(x₁, x₂)) = (1/2)x_1 + (4)x_2
POL(s) = 7/4
POL(0) = 4
POL(id) = 15/4

The value of delta used in the strict ordering is 7/8.
The following usable rules [17] were oriented:

app(add, 0) → id

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

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(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

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

POL(APP(x₁, x₂)) = (7/2)x_2
POL(cons) = 1/4
POL(map) = 0
POL(app(x₁, x₂)) = (1/4)x_1 + (7/2)x_2

The value of delta used in the strict ordering is 7/128.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

app(id, x) → x
app(add, 0) → id
app(app(add, app(s, x)), y) → app(s, app(app(add, x), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))

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.