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 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 PisEmptyProof (⇔)
        ↳24 TRUE

(0) Obligation:

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

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(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(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(cons, app(app(append, xs), ys))
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(append, xs), ys)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(append, xs)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)
APP(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(zip, xss)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(combine, app(app(zip, xs), ys))
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(zip, xs), ys)
APP(app(combine, xs), app(app(cons, ys), yss)) → APP(zip, xs)
APP(levels, app(app(node, x), xs)) → APP(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))
APP(levels, app(app(node, x), xs)) → APP(cons, app(app(cons, x), nil))
APP(levels, app(app(node, x), xs)) → APP(app(cons, x), nil)
APP(levels, app(app(node, x), xs)) → APP(cons, x)
APP(levels, app(app(node, x), xs)) → APP(app(combine, nil), app(app(map, levels), xs))
APP(levels, app(app(node, x), xs)) → APP(combine, nil)
APP(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)
APP(levels, app(app(node, x), xs)) → APP(map, levels)

The TRS R consists of the following rules:

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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 4 SCCs with 19 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

The TRS R consists of the following rules:

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
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(append) = 0   
POL(cons) = 0   

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(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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)

The TRS R consists of the following rules:

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), 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(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → APP(app(zip, xss), yss)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)

The TRS R consists of the following rules:

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


APP(app(combine, xs), app(app(cons, ys), yss)) → APP(app(combine, app(app(zip, xs), ys)), yss)
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(append) = 0   
POL(combine) = 0   
POL(cons) = 0   
POL(nil) = 0   
POL(zip) = 0   

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)

The TRS R consists of the following rules:

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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

(21) 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(levels, app(app(node, x), xs)) → APP(app(map, levels), xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

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(append, xs), nil) → xs
app(app(append, nil), ys) → ys
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))
app(app(zip, nil), yss) → yss
app(app(zip, xss), nil) → xss
app(app(zip, app(app(cons, xs), xss)), app(app(cons, ys), yss)) → app(app(cons, app(app(append, xs), ys)), app(app(zip, xss), yss))
app(app(combine, xs), nil) → xs
app(app(combine, xs), app(app(cons, ys), yss)) → app(app(combine, app(app(zip, xs), ys)), yss)
app(levels, app(app(node, x), xs)) → app(app(cons, app(app(cons, x), nil)), app(app(combine, nil), app(app(map, levels), xs)))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE