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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

APP(D, app(app(+, x), y)) → APP(app(+, app(D, x)), app(D, y))
APP(D, app(app(+, x), y)) → APP(+, app(D, x))
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(*, x), y)) → APP(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
APP(D, app(app(*, x), y)) → APP(+, app(app(*, y), app(D, x)))
APP(D, app(app(*, x), y)) → APP(app(*, y), app(D, x))
APP(D, app(app(*, x), y)) → APP(*, y)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(*, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(app(-, app(D, x)), app(D, y))
APP(D, app(app(-, x), y)) → APP(-, app(D, x))
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(minus, x)) → APP(minus, app(D, x))
APP(D, app(minus, x)) → APP(D, x)
APP(D, app(app(div, x), y)) → APP(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
APP(D, app(app(div, x), y)) → APP(-, app(app(div, app(D, x)), y))
APP(D, app(app(div, x), y)) → APP(app(div, app(D, x)), y)
APP(D, app(app(div, x), y)) → APP(div, app(D, x))
APP(D, app(app(div, x), y)) → APP(D, x)
APP(D, app(app(div, x), y)) → APP(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2))
APP(D, app(app(div, x), y)) → APP(div, app(app(*, x), app(D, y)))
APP(D, app(app(div, x), y)) → APP(app(*, x), app(D, y))
APP(D, app(app(div, x), y)) → APP(*, x)
APP(D, app(app(div, x), y)) → APP(D, y)
APP(D, app(app(div, x), y)) → APP(app(pow, y), 2)
APP(D, app(app(div, x), y)) → APP(pow, y)
APP(D, app(ln, x)) → APP(app(div, app(D, x)), x)
APP(D, app(ln, x)) → APP(div, app(D, x))
APP(D, app(ln, x)) → APP(D, x)
APP(D, app(app(pow, x), y)) → APP(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, y)))
APP(D, app(app(pow, x), y)) → APP(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x)))
APP(D, app(app(pow, x), y)) → APP(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))
APP(D, app(app(pow, x), y)) → APP(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1))))
APP(D, app(app(pow, x), y)) → APP(app(*, y), app(app(pow, x), app(app(-, y), 1)))
APP(D, app(app(pow, x), y)) → APP(*, y)
APP(D, app(app(pow, x), y)) → APP(app(pow, x), app(app(-, y), 1))
APP(D, app(app(pow, x), y)) → APP(app(-, y), 1)
APP(D, app(app(pow, x), y)) → APP(-, y)
APP(D, app(app(pow, x), y)) → APP(D, x)
APP(D, app(app(pow, x), y)) → APP(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, y))
APP(D, app(app(pow, x), y)) → APP(*, app(app(*, app(app(pow, x), y)), app(ln, x)))
APP(D, app(app(pow, x), y)) → APP(app(*, app(app(pow, x), y)), app(ln, x))
APP(D, app(app(pow, x), y)) → APP(*, app(app(pow, x), y))
APP(D, app(app(pow, x), y)) → APP(ln, x)
APP(D, app(app(pow, x), y)) → APP(D, y)
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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 46 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(minus, x)) → APP(D, x)
APP(D, app(app(div, x), y)) → APP(D, x)
APP(D, app(app(div, x), y)) → APP(D, y)
APP(D, app(ln, x)) → APP(D, x)
APP(D, app(app(pow, x), y)) → APP(D, x)
APP(D, app(app(pow, x), y)) → APP(D, y)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(8) 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(+(x, y)) → D1(y)
D1(+(x, y)) → D1(x)
D1(*(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(-(x, y)) → D1(x)
D1(-(x, y)) → D1(y)
D1(minus(x)) → D1(x)
D1(div(x, y)) → D1(x)
D1(div(x, y)) → D1(y)
D1(ln(x)) → D1(x)
D1(pow(x, y)) → D1(x)
D1(pow(x, y)) → D1(y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(D, app(app(+, x), y)) → APP(D, y)
APP(D, app(app(+, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, x)
APP(D, app(app(*, x), y)) → APP(D, y)
APP(D, app(app(-, x), y)) → APP(D, x)
APP(D, app(app(-, x), y)) → APP(D, y)
APP(D, app(app(div, x), y)) → APP(D, x)
APP(D, app(app(div, x), y)) → APP(D, y)
APP(D, app(app(pow, x), y)) → APP(D, x)
APP(D, app(app(pow, x), y)) → APP(D, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D1(x1)  =  x1
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x1, x2)
-(x1, x2)  =  -(x1, x2)
minus(x1)  =  x1
div(x1, x2)  =  div(x1, x2)
ln(x1)  =  x1
pow(x1, x2)  =  pow(x1, x2)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

APP(D, app(minus, x)) → APP(D, x)
APP(D, app(ln, x)) → APP(D, x)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(10) 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(minus(x)) → D1(x)
D1(ln(x)) → D1(x)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(D, app(ln, x)) → APP(D, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D1(x1)  =  x1
minus(x1)  =  x1
ln(x1)  =  ln(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

APP(D, app(minus, x)) → APP(D, x)

The TRS R consists of the following rules:

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(12) 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(minus(x)) → D1(x)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(D, app(minus, x)) → APP(D, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D1(x1)  =  x1
minus(x1)  =  minus(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
trivial

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

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

app(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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)  =  x2
app(x1, x2)  =  app(x1, x2)
cons  =  cons

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
trivial

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(D, t) → 1
app(D, constant) → 0
app(D, app(app(+, x), y)) → app(app(+, app(D, x)), app(D, y))
app(D, app(app(*, x), y)) → app(app(+, app(app(*, y), app(D, x))), app(app(*, x), app(D, y)))
app(D, app(app(-, x), y)) → app(app(-, app(D, x)), app(D, y))
app(D, app(minus, x)) → app(minus, app(D, x))
app(D, app(app(div, x), y)) → app(app(-, app(app(div, app(D, x)), y)), app(app(div, app(app(*, x), app(D, y))), app(app(pow, y), 2)))
app(D, app(ln, x)) → app(app(div, app(D, x)), x)
app(D, app(app(pow, x), y)) → app(app(+, app(app(*, app(app(*, y), app(app(pow, x), app(app(-, y), 1)))), app(D, x))), app(app(*, app(app(*, app(app(pow, x), y)), app(ln, x))), app(D, 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)

The set Q consists of the following terms:

app(D, t)
app(D, constant)
app(D, app(app(+, x0), x1))
app(D, app(app(*, x0), x1))
app(D, app(app(-, x0), x1))
app(D, app(minus, x0))
app(D, app(app(div, x0), x1))
app(D, app(ln, x0))
app(D, app(app(pow, x0), x1))
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(filter, x0), nil)
app(app(filter, x0), app(app(cons, x1), x2))
app(app(app(app(filter2, true), x0), x1), x2)
app(app(app(app(filter2, false), x0), x1), x2)

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