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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔)
        ↳11 TRUE
    ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 DependencyGraphProof (⇔)
        ↳16 TRUE

(0) Obligation:

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

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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(ack, 0), y) → APP(succ, y)
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(ack, app(succ, x)), y) → APP(ack, x)
APP(app(ack, app(succ, x)), y) → APP(succ, 0)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(ack, app(succ, x)), app(succ, y)) → APP(ack, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), 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(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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 2 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
succ > ack > [APP1, 0] > app1

Status:
APP1: [1]
app1: [1]
ack: []
succ: []
0: []


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
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)
ack  =  ack
succ  =  succ

Lexicographic path order with status [LPO].
Quasi-Precedence:
app2 > succ > ack

Status:
app2: [2,1]
ack: []
succ: []


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(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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.

(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(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(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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) 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 with status [LPO].
Quasi-Precedence:
map > [APP1, app2, true, false]
filter > [APP1, app2, true, false]
filter2 > [APP1, app2, true, false]

Status:
APP1: [1]
app2: [2,1]
map: []
cons: []
filter: []
filter2: []
true: []
false: []


The following usable rules [FROCOS05] were oriented: none

(14) 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(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, 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.

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE