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 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 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(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(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(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(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 19 less nodes.

(4) Complex Obligation (AND)

(5) 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)

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(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(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)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(*) = 1   
POL(+) = 1   
POL(-) = 0   
POL(APP(x1, x2)) = x2   
POL(D) = 0   
POL(app(x1, x2)) = 1 + x1 + x2   

The following usable rules [FROCOS05] were oriented: none

(7) 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(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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(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.

(11) 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(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 remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x1   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(false) = 1   
POL(filter) = 1   
POL(filter2) = 1   
POL(map) = 1   
POL(true) = 1   

The following usable rules [FROCOS05] were oriented: none

(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)

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(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)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x2   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(cons) = 0   
POL(map) = 0   

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(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(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) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE