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(id, x) -> x
app2(plus, 0) -> id
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(id, x) -> x
app2(plus, 0) -> id
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)

The TRS R consists of the following rules:

app2(id, x) -> x
app2(plus, 0) -> id
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)

The TRS R consists of the following rules:

app2(id, x) -> x
app2(plus, 0) -> id
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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 2 less nodes.

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

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

APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)

The TRS R consists of the following rules:

app2(id, x) -> x
app2(plus, 0) -> id
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(APP2(x1, x2)) = 2·x1   
POL(app2(x1, x2)) = 2 + 3·x1 + 3·x2   
POL(id) = 0   
POL(plus) = 3   
POL(s) = 0   

The following usable rules [14] were oriented:

app2(plus, 0) -> id



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

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

app2(id, x) -> x
app2(plus, 0) -> id
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))

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.