Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

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

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs)))
APP(app(filter, f), app(app(cons, x), xs)) → APP(if, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(cons, x), app(app(filter, f), xs))
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(if, true), x), y) → x
app(app(app(if, false), x), y) → y
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, 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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, 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.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
filter  =  filter
cons  =  cons

Recursive path order with status [RPO].
Precedence:

app₂ > filter

Status:

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

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

Q DP problem:
P is empty.
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(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(if, app(f, x)), app(app(cons, x), app(app(filter, f), xs))), app(app(filter, f), xs))

The set Q consists of the following terms:

app(app(app(if, true), x0), x1)
app(app(app(if, 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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE