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

(0) Obligation:

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

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

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(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, 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(plus, app(s, x)), y) → APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) → APP(plus, x)
APP(app(times, app(s, x)), y) → APP(app(plus, app(app(times, x), y)), y)
APP(app(times, app(s, x)), y) → APP(plus, app(app(times, x), y))
APP(app(times, app(s, x)), y) → APP(app(times, x), y)
APP(app(times, app(s, x)), y) → APP(times, x)
APP(inc, xs) → APP(app(map, app(plus, app(s, 0))), xs)
APP(inc, xs) → APP(map, app(plus, app(s, 0)))
APP(inc, xs) → APP(plus, app(s, 0))
APP(inc, xs) → APP(s, 0)
APP(double, xs) → APP(app(map, app(times, app(s, app(s, 0)))), xs)
APP(double, xs) → APP(map, app(times, app(s, app(s, 0))))
APP(double, xs) → APP(times, app(s, app(s, 0)))
APP(double, xs) → APP(s, app(s, 0))
APP(double, xs) → APP(s, 0)
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)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, 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 3 SCCs with 14 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(8) Obligation:

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

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

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(9) Obligation:

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

APP(inc, xs) → APP(app(map, app(plus, app(s, 0))), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(double, xs) → APP(app(map, app(times, app(s, app(s, 0)))), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(inc, xs) → APP(app(map, app(plus, app(s, 0))), xs)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
inc  =  inc
app(x1, x2)  =  x2
map  =  map
plus  =  plus
s  =  s
0  =  0
cons  =  cons
double  =  double
times  =  times
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

inc > [map, cons, nil] > [0, double]
inc > s > [plus, times] > [0, double]

Status:

cons: []
map: []
plus: []
times: []
inc: []
s: []
0: []
double: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

(11) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(double, xs) → APP(app(map, app(times, app(s, app(s, 0)))), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APP(double, xs) → APP(app(map, app(times, app(s, app(s, 0)))), xs)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
app(x1, x2)  =  x2
map  =  map
cons  =  cons
double  =  double
times  =  times
s  =  s
0  =  0
plus  =  plus
inc  =  inc
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:

cons > map > nil > [0, inc]
double > map > nil > [0, inc]
double > times > plus > [0, inc]
double > s > plus > [0, inc]

Status:

cons: []
map: []
plus: []
times: []
inc: []
s: []
double: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

(13) Obligation:

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

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)

The TRS R consists of the following rules:

app(app(plus, 0), y) → y
app(app(plus, app(s, x)), y) → app(s, app(app(plus, x), y))
app(app(times, 0), y) → 0
app(app(times, app(s, x)), y) → app(app(plus, app(app(times, x), y)), y)
app(inc, xs) → app(app(map, app(plus, app(s, 0))), xs)
app(double, xs) → app(app(map, app(times, app(s, app(s, 0)))), xs)
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))

The set Q consists of the following terms:

app(app(plus, 0), x0)
app(app(plus, app(s, x0)), x1)
app(app(times, 0), x0)
app(app(times, app(s, x0)), x1)
app(inc, x0)
app(double, x0)
app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))

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