/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 51 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) ATransformationProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QReductionProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) ATransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) ATransformationProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QReductionProof [EQUIVALENT, 0 ms]
        (31) QDP
        (32) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (33) YES
    (34) QDP
        (35) UsableRulesProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (38) YES


----------------------------------------

(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)
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)
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)
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) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(9)
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)

R is empty.
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) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   append1(cons(x, xs), ys) -> append1(xs, ys)

R is empty.
The set Q consists of the following terms:

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

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

(12) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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


----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   append1(cons(x, xs), ys) -> append1(xs, ys)

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

(14) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*append1(cons(x, xs), ys) -> append1(xs, ys)
The graph contains the following edges 1 > 1, 2 >= 2


----------------------------------------

(15)
YES

----------------------------------------

(16)
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.
----------------------------------------

(17) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(18)
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)

R is empty.
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.
----------------------------------------

(19) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   zip1(cons(xs, xss), cons(ys, yss)) -> zip1(xss, yss)

R is empty.
The set Q consists of the following terms:

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

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

(21) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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


----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   zip1(cons(xs, xss), cons(ys, yss)) -> zip1(xss, yss)

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

(23) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*zip1(cons(xs, xss), cons(ys, yss)) -> zip1(xss, yss)
The graph contains the following edges 1 > 1, 2 > 2


----------------------------------------

(24)
YES

----------------------------------------

(25)
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.
----------------------------------------

(26) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(27)
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(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))

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.
----------------------------------------

(28) ATransformationProof (EQUIVALENT)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem. 
----------------------------------------

(29)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   combine1(xs, cons(ys, yss)) -> combine1(zip(xs, ys), yss)

The TRS R consists of the following rules:

   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 set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
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(30) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   map(x0, nil)
   map(x0, cons(x1, x2))
   combine(x0, nil)
   combine(x0, cons(x1, x2))
   levels(node(x0, x1))


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(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   combine1(xs, cons(ys, yss)) -> combine1(zip(xs, ys), yss)

The TRS R consists of the following rules:

   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 set Q consists of the following terms:

   append(x0, nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   zip(nil, x0)
   zip(x0, nil)
   zip(cons(x0, x1), cons(x2, x3))

We have to consider all minimal (P,Q,R)-chains.
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(32) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*combine1(xs, cons(ys, yss)) -> combine1(zip(xs, ys), yss)
The graph contains the following edges 2 > 2


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(33)
YES

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(34)
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.
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(35) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
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(36)
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)

R is empty.
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.
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(37) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*APP(levels, app(app(node, x), xs)) -> APP(app(map, levels), xs)
The graph contains the following edges 2 > 2


*APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
The graph contains the following edges 1 > 1, 2 > 2


*APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
The graph contains the following edges 1 >= 1, 2 > 2


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(38)
YES