Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

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(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(app2(if, app2(p, x)), x)
APP2(app2(app2(until, p), f), x) -> APP2(if, app2(p, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

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(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(app2(if, app2(p, x)), x)
APP2(app2(app2(until, p), f), x) -> APP2(if, app2(p, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

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 3 less nodes.

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

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

APP2(app2(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
APP2(app2(app2(until, p), f), x) -> APP2(f, x)

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

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(app2(until, p), f), x) -> APP2(p, x)
APP2(app2(app2(until, p), f), x) -> APP2(f, x)
The remaining pairs can at least be oriented weakly.

APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 2·x1   
POL(app2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(false) = 0   
POL(if) = 0   
POL(true) = 0   
POL(until) = 0   

The following usable rules [14] were oriented: none



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

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

APP2(app2(app2(until, p), f), x) -> APP2(app2(app2(until, p), f), app2(f, x))

The TRS R consists of the following rules:

app2(app2(app2(if, true), x), y) -> x
app2(app2(app2(if, false), x), y) -> y
app2(app2(app2(until, p), f), x) -> app2(app2(app2(if, app2(p, x)), x), app2(app2(app2(until, p), f), app2(f, x)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.