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
  ↳22 QDP
    ↳23 QDPOrderProof (⇔)
    ↳24 QDP
    ↳25 PisEmptyProof (⇔)
    ↳26 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) 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(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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

(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(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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 19 less nodes.

(6) Complex Obligation (AND)

(7) 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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

append1(cons(x, xs), ys) → append1(xs, ys)

The a-transformed usable rules are
none


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: Combined order from the following AFS and order.
append1(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [RPO].
Precedence:
trivial


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(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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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(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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

zip1(cons(xs, xss), cons(ys, yss)) → zip1(xss, yss)

The a-transformed usable rules are
none


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: Combined order from the following AFS and order.
zip1(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [RPO].
Precedence:
trivial


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(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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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(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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

(18) 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

combine1(xs, cons(ys, yss)) → combine1(zip(xs, ys), yss)

The a-transformed usable rules are

zip(nil, yss) → yss
zip(xss, nil) → xss
zip(cons(xs, xss), cons(ys, yss)) → cons(append(xs, ys), zip(xss, yss))
append(xs, nil) → xs
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))


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: Combined order from the following AFS and order.
combine1(x1, x2)  =  combine1(x2)
cons(x1, x2)  =  cons(x1, x2)
zip(x1, x2)  =  zip(x1, x2)
nil  =  nil
append(x1, x2)  =  append(x1, x2)

Recursive Path Order [RPO].
Precedence:
[zip2, append2] > [combine11, cons2]


The following usable rules [FROCOS05] were oriented:

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(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))

(19) 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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

(22) 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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

(23) 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: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
levels  =  levels
node  =  node

Recursive Path Order [RPO].
Precedence:
map > app2
cons > app2
levels > app2
node > app2


The following usable rules [FROCOS05] were oriented: none

(24) 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)))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(app(append, x0), nil)
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)
app(app(zip, nil), x0)
app(app(zip, x0), nil)
app(app(zip, app(app(cons, x0), x1)), app(app(cons, x2), x3))
app(app(combine, x0), nil)
app(app(combine, x0), app(app(cons, x1), x2))
app(levels, app(app(node, x0), x1))

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE