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(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, 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(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)
APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(mapt, f), app(node, xs)) → APP(maptlist, f)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(mapt, f)
APP(app(mapt, f), app(node, xs)) → APP(node, app(app(maptlist, f), xs))
APP(app(mapt, f), app(leaf, x)) → APP(leaf, app(f, x))
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(cons, app(app(mapt, f), x))
APP(app(mapt, f), app(leaf, x)) → APP(f, x)

The TRS R consists of the following rules:

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, 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(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)
APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(mapt, f), app(node, xs)) → APP(maptlist, f)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(mapt, f)
APP(app(mapt, f), app(node, xs)) → APP(node, app(app(maptlist, f), xs))
APP(app(mapt, f), app(leaf, x)) → APP(leaf, app(f, x))
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(cons, app(app(mapt, f), x)), app(app(maptlist, f), xs))
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(cons, app(app(mapt, f), x))
APP(app(mapt, f), app(leaf, x)) → APP(f, x)

The TRS R consists of the following rules:

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, 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 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)
APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(mapt, f), app(leaf, x)) → APP(f, x)

The TRS R consists of the following rules:

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, 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(maptlist, f), app(app(cons, x), xs)) → APP(app(maptlist, f), xs)
APP(app(mapt, f), app(node, xs)) → APP(app(maptlist, f), xs)
APP(app(maptlist, f), app(app(cons, x), xs)) → APP(app(mapt, f), x)
APP(app(mapt, f), app(leaf, x)) → APP(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (1/2)x_1 + (1/4)x_2   
POL(cons) = 1/4   
POL(mapt) = 0   
POL(node) = 1/4   
POL(leaf) = 0   
POL(app(x1, x2)) = 1 + (1/4)x_1 + (4)x_2   
POL(maptlist) = 1/4   
The value of delta used in the strict ordering is 15/64.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

app(app(mapt, f), app(leaf, x)) → app(leaf, app(f, x))
app(app(mapt, f), app(node, xs)) → app(node, app(app(maptlist, f), xs))
app(app(maptlist, f), nil) → nil
app(app(maptlist, f), app(app(cons, x), xs)) → app(app(cons, app(app(mapt, f), x)), app(app(maptlist, 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.