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(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))

Q is empty.

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

APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(branch, app2(f, x))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(leaf, app2(f, x))

The TRS R consists of the following rules:

app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))

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:

APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(branch, app2(f, x))
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(leaf, app2(f, x))

The TRS R consists of the following rules:

app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

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

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

APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))

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


The following pairs can be oriented strictly and are deleted.


APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), r)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(app2(mapbt, f), l)
APP2(app2(mapbt, f), app2(leaf, x)) -> APP2(f, x)
APP2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = x1 + 3·x2   
POL(app2(x1, x2)) = 3 + 3·x1 + x2   
POL(branch) = 3   
POL(leaf) = 3   
POL(mapbt) = 2   

The following usable rules [14] 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:

app2(app2(mapbt, f), app2(leaf, x)) -> app2(leaf, app2(f, x))
app2(app2(mapbt, f), app2(app2(app2(branch, x), l), r)) -> app2(app2(app2(branch, app2(f, x)), app2(app2(mapbt, f), l)), app2(app2(mapbt, f), r))

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.