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:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

Q is empty.

The TRS is overlay and locally confluent. By [15] 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(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs))
APP(app(app(consif, true), x), ys) → APP(app(cons, x), ys)
APP(app(filter, f), app(app(cons, x), xs)) → APP(consif, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(consif, app(f, x)), x)
APP(app(app(consif, true), x), ys) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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

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

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs))
APP(app(app(consif, true), x), ys) → APP(app(cons, x), ys)
APP(app(filter, f), app(app(cons, x), xs)) → APP(consif, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(consif, app(f, x)), x)
APP(app(app(consif, true), x), ys) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

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

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

APP(app(app(consif, true), x), ys) → APP(app(cons, x), ys)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(consif, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(consif, app(f, x)), x)
APP(app(app(consif, true), x), ys) → APP(cons, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.

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

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

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.


APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
The remaining pairs can at least be oriented weakly.

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x1, x2)
filter  =  filter
cons  =  cons

Lexicographic path order with status [19].
Precedence:
app2 > APP1 > filter
cons > APP1 > filter

Status:
filter: multiset
APP1: [1]
app2: [1,2]
cons: multiset

The following usable rules [14] were oriented: none



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

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

FILTER(f, cons(x, xs)) → FILTER(f, xs)

R is empty.
The set Q consists of the following terms:

consif(true, x0, x1)
consif(false, x0, x1)
filter(x0, nil)
filter(x0, cons(x1, x2))

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


The following pairs can be oriented strictly and are deleted.


APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter, f), xs)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FILTER(x1, x2)  =  FILTER(x1, x2)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Precedence:
cons2 > FILTER2

Status:
FILTER2: [1,2]
cons2: multiset

The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

app(app(app(consif, true), x), ys) → app(app(cons, x), ys)
app(app(app(consif, false), x), ys) → ys
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(consif, app(f, x)), x), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(consif, true), x0), x1)
app(app(app(consif, false), x0), x1)
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.