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 Order [17,21] with Interpretation:

POL( plus ) = max{0, -3}


POL( 0 ) = 1


POL( APP2(x1, x2) ) = max{0, 2x1 + 3x2 - 3}


POL( s ) = 1


POL( app2(x1, x2) ) = 2x2 + 1


POL( id ) = 2



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.