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 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 TRUE

(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) 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

plus1(s(x), y) → plus1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(plus, app(s, x)), y) → APP(app(plus, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > plus12

Status:
s1: [1]
plus12: [2,1]


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(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) 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(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.

(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

times1(s(x), y) → times1(x, y)

The a-transformed usable rules are
none


The following pairs can be oriented strictly and are deleted.


APP(app(times, app(s, x)), y) → APP(app(times, x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > times12

Status:
s1: [1]
times12: [2,1]


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(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.

(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(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.

(18) 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)
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 remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
inc > APP2 > map > [app2, cons]
inc > APP2 > plus
inc > APP2 > s
inc > APP2 > 0
inc > APP2 > times
double > APP2 > map > [app2, cons]
double > APP2 > plus
double > APP2 > s
double > APP2 > 0
double > APP2 > times

Status:
APP2: [2,1]
cons: []
map: []
plus: []
times: []
inc: []
app2: [2,1]
s: []
0: []
double: []


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
P is empty.
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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE