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(, x), x) → e
app(app(, e), x) → x
app(app(, x), app(app(., x), y)) → y
app(app(, app(app(/, x), y)), x) → y
app(app(/, x), x) → e
app(app(/, x), e) → x
app(app(/, app(app(., y), x)), x) → y
app(app(/, x), app(app(, y), x)) → y
app(app(., e), x) → x
app(app(., x), e) → x
app(app(., x), app(app(, x), y)) → y
app(app(., app(app(/, y), x)), x) → y
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)

The TRS R consists of the following rules:

app(app(, x), x) → e
app(app(, e), x) → x
app(app(, x), app(app(., x), y)) → y
app(app(, app(app(/, x), y)), x) → y
app(app(/, x), x) → e
app(app(/, x), e) → x
app(app(/, app(app(., y), x)), x) → y
app(app(/, x), app(app(, y), x)) → y
app(app(., e), x) → x
app(app(., x), e) → x
app(app(., x), app(app(, x), y)) → y
app(app(., app(app(/, y), x)), x) → y
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 9 less nodes.

(4) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(, x), x) → e
app(app(, e), x) → x
app(app(, x), app(app(., x), y)) → y
app(app(, app(app(/, x), y)), x) → y
app(app(/, x), x) → e
app(app(/, x), e) → x
app(app(/, app(app(., y), x)), x) → y
app(app(/, x), app(app(, y), x)) → y
app(app(., e), x) → x
app(app(., x), e) → x
app(app(., x), app(app(, x), y)) → y
app(app(., app(app(/, y), x)), x) → y
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

(5) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false
app(x1, x2)  =  app(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 > 2

• APP(app(map, f), app(app(cons, x), xs)) → APP(f, x) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 > 2

• APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

• APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(6) TRUE