Termination w.r.t. Q proof of /tmp/tmpkjw3NB/Ex6Folding.trs

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(fold, f), nil), x) → x
app(app(app(fold, f), app(app(cons, h), t)), x) → app(app(app(fold, f), t), app(app(f, x), h))
app(sum, l) → app(app(app(fold, add), l), 0)
app(app(app(fold, mul), l), 1) → app(prod, l)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

app(app(app(fold, f), nil), x) → x
app(app(app(fold, f), app(app(cons, h), t)), x) → app(app(app(fold, f), t), app(app(f, x), h))
app(sum, l) → app(app(app(fold, add), l), 0)
app(app(app(fold, mul), l), 1) → app(prod, l)

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(fold, f), app(app(cons, h), t)), x) → APP(f, x)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(f, x), h)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(app(fold, f), t), app(app(f, x), h))
APP(sum, l) → APP(fold, add)
APP(sum, l) → APP(app(fold, add), l)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(fold, f), t)
APP(app(app(fold, mul), l), 1) → APP(prod, l)
APP(sum, l) → APP(app(app(fold, add), l), 0)

The TRS R consists of the following rules:

app(app(app(fold, f), nil), x) → x
app(app(app(fold, f), app(app(cons, h), t)), x) → app(app(app(fold, f), t), app(app(f, x), h))
app(sum, l) → app(app(app(fold, add), l), 0)
app(app(app(fold, mul), l), 1) → app(prod, l)

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(fold, f), app(app(cons, h), t)), x) → APP(f, x)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(f, x), h)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(app(fold, f), t), app(app(f, x), h))
APP(sum, l) → APP(fold, add)
APP(sum, l) → APP(app(fold, add), l)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(fold, f), t)
APP(app(app(fold, mul), l), 1) → APP(prod, l)
APP(sum, l) → APP(app(app(fold, add), l), 0)

The TRS R consists of the following rules:

app(app(app(fold, f), nil), x) → x
app(app(app(fold, f), app(app(cons, h), t)), x) → app(app(app(fold, f), t), app(app(f, x), h))
app(sum, l) → app(app(app(fold, add), l), 0)
app(app(app(fold, mul), l), 1) → app(prod, l)

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(f, x)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(f, x), h)
APP(app(app(fold, f), app(app(cons, h), t)), x) → APP(app(app(fold, f), t), app(app(f, x), h))
APP(sum, l) → APP(app(app(fold, add), l), 0)

The TRS R consists of the following rules:

app(app(app(fold, f), nil), x) → x
app(app(app(fold, f), app(app(cons, h), t)), x) → app(app(app(fold, f), t), app(app(f, x), h))
app(sum, l) → app(app(app(fold, add), l), 0)
app(app(app(fold, mul), l), 1) → app(prod, l)

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