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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE

(0) Obligation:

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

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)
APP(app(app(f, 0), 1), x) → APP(app(f, app(s, x)), x)
APP(app(app(f, 0), 1), x) → APP(f, app(s, x))
APP(app(app(f, 0), 1), x) → APP(s, x)
APP(app(app(f, x), y), app(s, z)) → APP(s, app(app(app(f, 0), 1), z))
APP(app(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)
APP(app(app(f, x), y), app(s, z)) → APP(app(f, 0), 1)
APP(app(app(f, x), y), app(s, z)) → APP(f, 0)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(cons, app(fun, x)), app(app(map, fun), xs))
APP(app(map, fun), app(app(cons, x), xs)) → APP(cons, app(fun, x))
APP(app(map, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(map, fun), app(app(cons, x), xs)) → APP(app(map, fun), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(app(filter2, app(fun, x)), fun), x)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(app(filter2, app(fun, x)), fun)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(filter2, app(fun, x))
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(cons, x), app(app(filter, fun), xs))
APP(app(app(app(filter2, true), fun), x), xs) → APP(cons, x)
APP(app(app(app(filter2, true), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, true), fun), x), xs) → APP(filter, fun)
APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), xs)
APP(app(app(app(filter2, false), fun), x), xs) → APP(filter, fun)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), 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 2 SCCs with 16 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z)
APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x)

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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

(6) 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:
f  =  f
s  =  s
0  =  0
1  =  1
app(x1, x2)  =  app(x2)

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

  • APP(app(app(f, x), y), app(s, z)) → APP(app(app(f, 0), 1), z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 2 > 2

  • APP(app(app(f, 0), 1), x) → APP(app(app(f, app(s, x)), x), x) (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

(7) TRUE

(8) Obligation:

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

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

The TRS R consists of the following rules:

app(app(app(f, 0), 1), x) → app(app(app(f, app(s, x)), x), x)
app(app(app(f, x), y), app(s, z)) → app(s, app(app(app(f, 0), 1), z))
app(app(map, fun), nil) → nil
app(app(map, fun), app(app(cons, x), xs)) → app(app(cons, app(fun, x)), app(app(map, fun), xs))
app(app(filter, fun), nil) → nil
app(app(filter, fun), app(app(cons, x), xs)) → app(app(app(app(filter2, app(fun, x)), fun), x), xs)
app(app(app(app(filter2, true), fun), x), xs) → app(app(cons, x), app(app(filter, fun), xs))
app(app(app(app(filter2, false), fun), x), xs) → app(app(filter, fun), xs)

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

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

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

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

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

  • APP(app(app(app(filter2, false), fun), x), xs) → APP(app(filter, fun), 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

(10) TRUE