YES Termination w.r.t. Q proof of Applicative_05_TreeLevels.ari

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

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

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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.

(30) QReductionProof (EQUIVALENT transformation)

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))

(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.

(32) QDPSizeChangeProof (EQUIVALENT transformation)

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

(33) YES

(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.

(35) UsableRulesProof (EQUIVALENT transformation)

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.

(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.

(37) QDPSizeChangeProof (EQUIVALENT transformation)

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

(38) YES