Termination w.r.t. Q proof of /tmp/tmp1n5zWs/ReverseLastInit.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(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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(reverse2, app(app(cons, x), xs)), l) → APP(reverse2, xs)
APP(reverse, l) → APP(reverse2, l)
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(cons, x), l)
LAST → APP(compose, hd)
APP(reverse, l) → APP(app(reverse2, l), nil)
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))
INIT → APP(compose, reverse)
APP(app(app(compose, f), g), x) → APP(g, app(f, x))
INIT → APP(app(compose, tl), reverse)
INIT → APP(compose, tl)
APP(app(app(compose, f), g), x) → APP(f, x)
INIT → APP(app(compose, reverse), app(app(compose, tl), reverse))
LAST → APP(app(compose, hd), reverse)

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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(reverse2, app(app(cons, x), xs)), l) → APP(reverse2, xs)
APP(reverse, l) → APP(reverse2, l)
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(cons, x), l)
LAST → APP(compose, hd)
APP(reverse, l) → APP(app(reverse2, l), nil)
APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))
INIT → APP(compose, reverse)
APP(app(app(compose, f), g), x) → APP(g, app(f, x))
INIT → APP(app(compose, tl), reverse)
INIT → APP(compose, tl)
APP(app(app(compose, f), g), x) → APP(f, x)
INIT → APP(app(compose, reverse), app(app(compose, tl), reverse))
LAST → APP(app(compose, hd), reverse)

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

APP(app(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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(reverse2, app(app(cons, x), xs)), l) → APP(app(reverse2, xs), app(app(cons, x), l))

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

POL(APP(x₁, x₂)) = (2)x_1
POL(reverse2) = 0
POL(cons) = 0
POL(app(x₁, x₂)) = 1 + x_1 + x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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

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

APP(app(app(compose, f), g), x) → APP(g, app(f, x))
APP(app(app(compose, f), g), x) → APP(f, x)

The TRS R consists of the following rules:

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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(app(compose, f), g), x) → APP(g, app(f, x))
APP(app(app(compose, f), g), x) → APP(f, x)

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

POL(APP(x₁, x₂)) = (3)x_1
POL(reverse2) = 3
POL(cons) = 1
POL(tl) = 0
POL(reverse) = 2
POL(compose) = 4
POL(app(x₁, x₂)) = 2 + x_1 + (2)x_2
POL(nil) = 3
POL(hd) = 0

The value of delta used in the strict ordering is 24.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

app(app(app(compose, f), g), x) → app(g, app(f, x))
app(reverse, l) → app(app(reverse2, l), nil)
app(app(reverse2, nil), l) → l
app(app(reverse2, app(app(cons, x), xs)), l) → app(app(reverse2, xs), app(app(cons, x), l))
app(hd, app(app(cons, x), xs)) → x
app(tl, app(app(cons, x), xs)) → xs
last → app(app(compose, hd), reverse)
init → app(app(compose, reverse), app(app(compose, tl), reverse))

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.