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:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

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(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))
APP(app(app(until, p), f), x) → APP(f, x)
APP(app(app(until, p), f), x) → APP(p, x)
APP(app(app(until, p), f), x) → APP(app(if, app(p, x)), x)
APP(app(app(until, p), f), x) → APP(if, app(p, x))
APP(app(app(until, p), f), x) → APP(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(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:

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))
APP(app(app(until, p), f), x) → APP(f, x)
APP(app(app(until, p), f), x) → APP(p, x)
APP(app(app(until, p), f), x) → APP(app(if, app(p, x)), x)
APP(app(app(until, p), f), x) → APP(if, app(p, x))
APP(app(app(until, p), f), x) → APP(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

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 3 less nodes.

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

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

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))
APP(app(app(until, p), f), x) → APP(f, x)
APP(app(app(until, p), f), x) → APP(p, x)

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

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(app(until, p), f), x) → APP(f, x)
APP(app(app(until, p), f), x) → APP(p, x)
The remaining pairs can at least be oriented weakly.

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))
Used ordering: Polynomial interpretation [25,35]:

POL(until) = 4   
POL(APP(x1, x2)) = (4)x_1   
POL(if) = 3   
POL(true) = 0   
POL(false) = 0   
POL(app(x1, x2)) = 1 + (4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 276.
The following usable rules [17] were oriented: none



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

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

APP(app(app(until, p), f), x) → APP(app(app(until, p), f), app(f, x))

The TRS R consists of the following rules:

app(app(app(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(app(until, p), f), x) → app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x)))

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