(0) Obligation:

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

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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(f, app(s, x)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, x), y))
APP(app(g, x), app(c, y)) → APP(app(g, x), y)
APP(app(g, x), app(c, y)) → APP(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
APP(app(g, x), app(c, y)) → APP(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y))
APP(app(g, x), app(c, y)) → APP(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y)))
APP(app(g, x), app(c, y)) → APP(if, app(f, x))
APP(app(g, x), app(c, y)) → APP(f, x)
APP(app(g, x), app(c, y)) → APP(c, app(app(g, app(s, x)), y))
APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
APP(app(g, x), app(c, y)) → APP(g, app(s, x))
APP(app(g, x), app(c, y)) → APP(s, x)
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(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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 3 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(f, app(s, x)) → APP(f, x)

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(f, app(s, x)) → 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)  =  APP(x2)
f  =  f
app(x1, x2)  =  app(x2)
s  =  s

Recursive path order with status [RPO].
Quasi-Precedence:
[f, s] > [APP1, app1]

Status:
APP1: multiset
f: multiset
app1: multiset
s: multiset


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
APP(app(g, x), app(c, y)) → APP(app(g, x), y)

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(g, x), app(c, y)) → APP(app(g, app(s, x)), y)
APP(app(g, x), app(c, y)) → APP(app(g, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1, x2)
app(x1, x2)  =  app(x2)
g  =  g
c  =  c
s  =  s

Recursive path order with status [RPO].
Quasi-Precedence:
g > [APP2, s] > app1
c > app1

Status:
APP2: [2,1]
app1: multiset
g: multiset
c: multiset
s: multiset


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

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

The TRS R consists of the following rules:

app(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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(app(app(app(filter2, app(fun, x)), fun), x), xs)
APP(app(filter, fun), app(app(cons, x), xs)) → APP(fun, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
filter  =  filter
filter2  =  filter2
true  =  true
false  =  false
f  =  f
0  =  0
1  =  1
s  =  s
if  =  if
g  =  g
c  =  c
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
filter2 > cons > [app2, map, true] > [f, s, g] > [false, 1] > APP1 > filter
filter2 > cons > [app2, map, true] > [f, s, g] > if > filter
filter2 > cons > [app2, map, true] > [f, s, g] > c > filter
filter2 > cons > [app2, map, true] > nil > filter
0 > filter

Status:
APP1: multiset
app2: multiset
map: multiset
cons: multiset
filter: multiset
filter2: multiset
true: multiset
false: multiset
f: multiset
0: multiset
1: multiset
s: multiset
if: multiset
g: multiset
c: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

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(f, 0) → true
app(f, 1) → false
app(f, app(s, x)) → app(f, x)
app(app(app(if, true), app(s, x)), app(s, y)) → app(s, x)
app(app(app(if, false), app(s, x)), app(s, y)) → app(s, y)
app(app(g, x), app(c, y)) → app(c, app(app(g, x), y))
app(app(g, x), app(c, y)) → app(app(g, x), app(app(app(if, app(f, x)), app(c, app(app(g, app(s, x)), y))), app(c, y)))
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.

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(19) TRUE