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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
  ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 DependencyGraphProof (⇔)
    ↳22 TRUE

(0) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) → APP(le, x)
APP(app(minus, x), app(s, y)) → APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(if_gcd, app(app(le, y), x)), app(s, x))
APP(app(gcd, app(s, x)), app(s, y)) → APP(if_gcd, app(app(le, y), x))
APP(app(gcd, app(s, x)), app(s, y)) → APP(app(le, y), x)
APP(app(gcd, app(s, x)), app(s, y)) → APP(le, y)
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(gcd, app(app(minus, x), y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(minus, x), y)
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(minus, x)
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(gcd, app(app(minus, y), x))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(minus, y), x)
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(minus, 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(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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 4 SCCs with 21 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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

minus1(x, s(y)) → minus1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(minus, x), app(s, y)) → APP(app(minus, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
minus1(x1, x2)  =  minus1(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > minus11

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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.

(13) 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

le1(s(x), s(y)) → le1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(le, app(s, x)), app(s, y)) → APP(app(le, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
le1(x1, x2)  =  le1(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
s1 > le11

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

APP(app(gcd, app(s, x)), app(s, y)) → APP(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
APP(app(app(if_gcd, true), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, x), y)), app(s, y))
APP(app(app(if_gcd, false), app(s, x)), app(s, y)) → APP(app(gcd, app(app(minus, y), x)), app(s, x))

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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.

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

The TRS R consists of the following rules:

app(app(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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) 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(app(app(app(filter2, app(f, x)), f), x), xs)
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
le  =  le
0  =  0
s  =  s
pred  =  pred
minus  =  minus
gcd  =  gcd
if_gcd  =  if_gcd
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:
filter > APP1 > app2
filter > APP1 > map
filter > APP1 > filter2
filter > nil
true > APP1 > app2
true > APP1 > map
true > APP1 > filter2
true > cons
true > minus > app2
true > gcd > app2
true > gcd > s
true > gcd > ifgcd
false > APP1 > app2
false > APP1 > map
false > APP1 > filter2
false > minus > app2
false > gcd > app2
false > gcd > s
false > gcd > ifgcd
le > app2

The following usable rules [FROCOS05] were oriented: none

(20) 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(le, 0), y) → true
app(app(le, app(s, x)), 0) → false
app(app(le, app(s, x)), app(s, y)) → app(app(le, x), y)
app(pred, app(s, x)) → x
app(app(minus, x), 0) → x
app(app(minus, x), app(s, y)) → app(pred, app(app(minus, x), y))
app(app(gcd, 0), y) → y
app(app(gcd, app(s, x)), 0) → app(s, x)
app(app(gcd, app(s, x)), app(s, y)) → app(app(app(if_gcd, app(app(le, y), x)), app(s, x)), app(s, y))
app(app(app(if_gcd, true), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, x), y)), app(s, y))
app(app(app(if_gcd, false), app(s, x)), app(s, y)) → app(app(gcd, app(app(minus, y), x)), app(s, x))
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(app(le, 0), x0)
app(app(le, app(s, x0)), 0)
app(app(le, app(s, x0)), app(s, x1))
app(pred, app(s, x0))
app(app(minus, x0), 0)
app(app(minus, x0), app(s, x1))
app(app(gcd, 0), x0)
app(app(gcd, app(s, x0)), 0)
app(app(gcd, app(s, x0)), app(s, x1))
app(app(app(if_gcd, true), app(s, x0)), app(s, x1))
app(app(app(if_gcd, false), app(s, x0)), app(s, 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.

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) TRUE