Termination w.r.t. Q of the given QTRS could not be shown:

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 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 DependencyGraphProof (⇔)
    ↳20 TRUE

(0) Obligation:

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

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))
APP(app(h, z), app(e, x)) → APP(h, app(c, z))
APP(app(h, z), app(e, x)) → APP(c, z)
APP(app(h, z), app(e, x)) → APP(app(d, z), x)
APP(app(h, z), app(e, x)) → APP(d, z)
APP(app(d, z), app(app(g, 0), 0)) → APP(e, 0)
APP(app(d, z), app(app(g, x), y)) → APP(app(g, app(e, x)), app(app(d, z), y))
APP(app(d, z), app(app(g, x), y)) → APP(g, app(e, x))
APP(app(d, z), app(app(g, x), y)) → APP(e, x)
APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(g, app(app(d, app(c, z)), app(app(g, x), y)))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(d, z)
APP(app(g, app(e, x)), app(e, y)) → APP(e, app(app(g, x), y))
APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)
APP(app(g, app(e, x)), app(e, y)) → APP(g, x)
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(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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 4 SCCs with 23 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(g, app(e, x)), app(e, y)) → APP(app(g, x), y)

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

(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

g1(e(x), e(y)) → g1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(g, app(e, x)), app(e, 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)  =  x2
e(x1)  =  e(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(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

(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(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

(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

d1(c(z), g(g(x, y), 0)) → d1(c(z), g(x, y))
d1(z, g(x, y)) → d1(z, y)
d1(c(z), g(g(x, y), 0)) → d1(z, g(x, y))

The a-transformed usable rules are

g(e(x), e(y)) → e(g(x, y))


The following pairs can be oriented strictly and are deleted.


APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, app(c, z)), app(app(g, x), y))
APP(app(d, z), app(app(g, x), y)) → APP(app(d, z), y)
APP(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → APP(app(d, z), app(app(g, x), y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
d1(x1, x2)  =  x2
c(x1)  =  x1
g(x1, x2)  =  g(x1, x2)
0  =  0
e(x1)  =  e(x1)

Lexicographic Path Order [LPO].
Precedence:
0 > g2 > e1


The following usable rules [FROCOS05] were oriented:

app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, x), y))

(12) Obligation:

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

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

(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(h, z), app(e, x)) → APP(app(h, app(c, z)), app(app(d, z), x))

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

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

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
map > [APP1, app2, cons, filter2, true]
filter > [APP1, app2, cons, filter2, true]
false > [APP1, app2, cons, filter2, true]


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

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

The TRS R consists of the following rules:

app(app(h, z), app(e, x)) → app(app(h, app(c, z)), app(app(d, z), x))
app(app(d, z), app(app(g, 0), 0)) → app(e, 0)
app(app(d, z), app(app(g, x), y)) → app(app(g, app(e, x)), app(app(d, z), y))
app(app(d, app(c, z)), app(app(g, app(app(g, x), y)), 0)) → app(app(g, app(app(d, app(c, z)), app(app(g, x), y))), app(app(d, z), app(app(g, x), y)))
app(app(g, app(e, x)), app(e, y)) → app(e, app(app(g, 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.

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) TRUE