Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

The set Q consists of the following terms:

app(app(iterate, x0), x1)


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(iterate, f), x) → APP(app(iterate, f), app(f, x))
APP(app(iterate, f), x) → APP(f, x)
APP(app(iterate, f), x) → APP(cons, x)
APP(app(iterate, f), x) → APP(app(cons, x), app(app(iterate, f), app(f, x)))

The TRS R consists of the following rules:

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

The set Q consists of the following terms:

app(app(iterate, x0), x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

APP(app(iterate, f), x) → APP(app(iterate, f), app(f, x))
APP(app(iterate, f), x) → APP(f, x)
APP(app(iterate, f), x) → APP(cons, x)
APP(app(iterate, f), x) → APP(app(cons, x), app(app(iterate, f), app(f, x)))

The TRS R consists of the following rules:

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

The set Q consists of the following terms:

app(app(iterate, x0), x1)

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

APP(app(iterate, f), x) → APP(app(iterate, f), app(f, x))
APP(app(iterate, f), x) → APP(f, x)

The TRS R consists of the following rules:

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

The set Q consists of the following terms:

app(app(iterate, x0), x1)

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

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

POL(APP(x1, x2)) = x1   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(iterate) = 1   

The following usable rules [17] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ MNOCProof

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

APP(app(iterate, f), x) → APP(app(iterate, f), app(f, x))

The TRS R consists of the following rules:

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

The set Q consists of the following terms:

app(app(iterate, x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ MNOCProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

APP(app(iterate, f), x) → APP(app(iterate, f), app(f, x))

The TRS R consists of the following rules:

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))

Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP(app(iterate, f), x) → APP(app(iterate, f), app(f, x))

The TRS R consists of the following rules:

app(app(iterate, f), x) → app(app(cons, x), app(app(iterate, f), app(f, x)))


s = APP(app(iterate, f), x) evaluates to t =APP(app(iterate, f), app(f, x))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP(app(iterate, f), x) to APP(app(iterate, f), app(f, x)).