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:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LAST -> APP2(app2(compose, hd), reverse)
LAST -> APP2(compose, hd)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(cons, x), l)
INIT -> APP2(app2(compose, tl), reverse)
INIT -> APP2(compose, tl)
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(reverse2, xs)
INIT -> APP2(compose, reverse)
APP2(reverse, l) -> APP2(reverse2, l)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(reverse, l) -> APP2(app2(reverse2, l), nil)
INIT -> APP2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

LAST -> APP2(app2(compose, hd), reverse)
LAST -> APP2(compose, hd)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(cons, x), l)
INIT -> APP2(app2(compose, tl), reverse)
INIT -> APP2(compose, tl)
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(reverse2, xs)
INIT -> APP2(compose, reverse)
APP2(reverse, l) -> APP2(reverse2, l)
APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(reverse, l) -> APP2(app2(reverse2, l), nil)
INIT -> APP2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 10 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(reverse2, app2(app2(cons, x), xs)), l) -> APP2(app2(reverse2, xs), app2(app2(cons, x), l))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app1(x2)
reverse2  =  reverse2
cons  =  cons

Lexicographic Path Order [19].
Precedence:
cons > [app1, reverse2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(app2(app2(compose, f), g), x) -> APP2(g, app2(f, x))
APP2(app2(app2(compose, f), g), x) -> APP2(f, x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
app2(x1, x2)  =  app2(x1, x2)
compose  =  compose
reverse  =  reverse
reverse2  =  reverse2
nil  =  nil
tl  =  tl
cons  =  cons
hd  =  hd

Lexicographic Path Order [19].
Precedence:
app2 > APP1
[reverse, nil]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

app2(app2(app2(compose, f), g), x) -> app2(g, app2(f, x))
app2(reverse, l) -> app2(app2(reverse2, l), nil)
app2(app2(reverse2, nil), l) -> l
app2(app2(reverse2, app2(app2(cons, x), xs)), l) -> app2(app2(reverse2, xs), app2(app2(cons, x), l))
app2(hd, app2(app2(cons, x), xs)) -> x
app2(tl, app2(app2(cons, x), xs)) -> xs
last -> app2(app2(compose, hd), reverse)
init -> app2(app2(compose, reverse), app2(app2(compose, tl), reverse))

The set Q consists of the following terms:

app2(app2(app2(compose, x0), x1), x2)
app2(reverse, x0)
app2(app2(reverse2, nil), x0)
app2(app2(reverse2, app2(app2(cons, x0), x1)), x2)
app2(hd, app2(app2(cons, x0), x1))
app2(tl, app2(app2(cons, x0), x1))
last
init

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