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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 DependencyGraphProof (⇔)
    ↳19 TRUE

(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]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

f1(s(x)) → f1(x)

The a-transformed usable rules are
none


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.
f1(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


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]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

g1(x, c(y)) → g1(s(x), y)
g1(x, c(y)) → g1(x, y)

The a-transformed usable rules are
none


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.
g1(x1, x2)  =  g1(x2)
c(x1)  =  c(x1)
s(x1)  =  s

Lexicographic Path Order [LPO].
Precedence:
[g11, c1] > s


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

Lexicographic Path Order [LPO].
Precedence:
[filter, true, f, 0, if, c] > [APP1, app2, map, cons, filter2, false, 1, s, g, nil]


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