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(id, x) → x
app(plus, 0) → id
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(id, x) → x
app(plus, 0) → id
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(plus, x)

The TRS R consists of the following rules:

app(id, x) → x
app(plus, 0) → id
app(app(plus, app(s, x)), y) → app(s, app(app(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:

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(plus, x)

The TRS R consists of the following rules:

app(id, x) → x
app(plus, 0) → id
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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 2 less nodes.

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

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

APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)

The TRS R consists of the following rules:

app(id, x) → x
app(plus, 0) → id
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))

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(plus, app(s, x)), y) → APP(app(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x1, x2)) = (2)x_1   
POL(plus) = 1   
POL(app(x1, x2)) = (2)x_1 + x_2   
POL(s) = 4   
POL(0) = 0   
POL(id) = 2   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

app(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:

app(id, x) → x
app(plus, 0) → id
app(app(plus, app(s, x)), y) → app(s, app(app(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.