YES Termination proof of AProVE_24_combination.ari

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 RelADPP
        ↳6 RelADPCleverAfsProof (⇒, 80 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 RelADPP
        ↳13 RelADPCleverAfsProof (⇒, 75 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔, 0 ms)
        ↳18 YES
    ↳19 RelADPP
        ↳20 RelADPCleverAfsProof (⇒, 84 ms)
        ↳21 QDP
        ↳22 MRRProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 MRRProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 MRRProof (⇔, 8 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 RelADPP
        ↳31 RelADPCleverAfsProof (⇒, 82 ms)
        ↳32 QDP
        ↳33 MRRProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 PisEmptyProof (⇔, 0 ms)
        ↳36 YES
    ↳37 RelADPP
        ↳38 RelADPDerelatifyingProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QDPOrderProof (⇔, 25 ms)
        ↳41 QDP
        ↳42 PisEmptyProof (⇔, 0 ms)
        ↳43 YES
    ↳44 RelADPP
        ↳45 RelADPDerelatifyingProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QDPOrderProof (⇔, 20 ms)
        ↳48 QDP
        ↳49 PisEmptyProof (⇔, 0 ms)
        ↳50 YES
    ↳51 RelADPP
        ↳52 RelADPReductionPairProof (⇔, 53 ms)
        ↳53 RelADPP
        ↳54 DAbsisEmptyProof (⇔, 0 ms)
        ↳55 YES

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
divL(x, nil) → x
divL(x, cons(y, xs)) → divL(div(x, y), xs)
app(nil, y) → y
app(cons(n, x), y) → cons(n, app(x, y))
reverse(nil) → nil
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
shuffle(nil) → nil
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

The relative TRS consists of the following S rules:

divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].

(2) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, 0) → x
minus(s(x), s(y)) → MINUS(x, y)
div(0, s(y)) → 0
div(s(x), s(y)) → s(DIV(minus(x, y), s(y)))
div(s(x), s(y)) → s(div(MINUS(x, y), s(y)))
divL(x, nil) → x
divL(x, cons(y, xs)) → DIVL(div(x, y), xs)
divL(x, cons(y, xs)) → divL(DIV(x, y), xs)
app(nil, y) → y
app(cons(n, x), y) → cons(n, APP(x, y))
reverse(nil) → nil
reverse(cons(n, x)) → APP(reverse(x), cons(n, nil))
reverse(cons(n, x)) → app(REVERSE(x), cons(n, nil))
shuffle(nil) → nil
shuffle(cons(n, x)) → cons(n, SHUFFLE(reverse(x)))
shuffle(cons(n, x)) → cons(n, shuffle(REVERSE(x)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, CONCAT(v, y))
less_leaves(x, leaf) → false
less_leaves(leaf, cons(w, z)) → true
less_leaves(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
less_leaves(cons(u, v), cons(w, z)) → less_leaves(CONCAT(u, v), concat(w, z))
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), CONCAT(w, z))

and relative ADPs:

divL(z, cons(x, cons(y, xs))) → DIVL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → SHUFFLE(cons(m, cons(n, x)))

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
7 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 7 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(s(x), s(y)) → MINUS(x, y)

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(6) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:MINUS_2 = 0 true = less_leaves_2 = app_2 = divL_2 = concat_2 = leaf = xs = shuffle_1 = nil = false = s_1 = reverse_1 = div_2 = 0 = minus_2 = 1 cons_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)
div(x1, x2)  =  div(x1, x2)
minus(x1, x2)  =  x1
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

MINUS1 > s1
div2 > s1
0 > s1

Status:
MINUS1: [1]
s1: multiset
div2: [2,1]
0: multiset

(7) Obligation:

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

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

div0(00, s0(y)) → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(x, cons0(y, xs0)) → divL0(div0(x, y), xs0)
shuffle0(nil0) → nil0
divL0(x, nil0) → x
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
less_leaves0(leaf0, cons0(w, z)) → true0
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

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

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MINUS(s0(y)) → MINUS(y)

Strictly oriented rules of the TRS R:

div0(00, s0(y)) → 00
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(x, cons0(y, xs0)) → divL0(div0(x, y), xs0)
divL0(x, nil0) → x
reverse0(nil0) → nil0
less_leaves0(leaf0, cons0(w, z)) → true0
minus(s0(x)) → minus(x)
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = 2·x1   
POL(app0(x1, x2)) = x1 + x2   
POL(concat0(x1, x2)) = 2·x1 + x2   
POL(cons0(x1, x2)) = 2 + 2·x1 + x2   
POL(div0(x1, x2)) = x1 + x2   
POL(divL0(x1, x2)) = 2 + x1 + x2   
POL(false0) = 2   
POL(leaf0) = 0   
POL(less_leaves0(x1, x2)) = 2 + 2·x1 + x2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(reverse0(x1)) = 1 + x1   
POL(s0(x1)) = 2 + x1   
POL(shuffle0(x1)) = 2·x1   
POL(true0) = 0   
POL(xs0) = 2   

(9) Obligation:

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

divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
concat0(leaf0, y) → y
shuffle0(nil0) → nil0
less_leaves0(x, leaf0) → false0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(x) → x

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) YES

(12) Obligation:

Relative ADP Problem with
absolute ADPs:

app(cons(n, x), y) → cons(n, APP(x, y))

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(13) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:true = less_leaves_2 = app_2 = divL_2 = concat_2 = leaf = xs = shuffle_1 = APP_2 = nil = false = s_1 = reverse_1 = div_2 = 0, 1 0 = minus_2 = 1 cons_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
div(x1, x2)  =  div
s(x1)  =  x1
minus(x1, x2)  =  x1
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

cons2 > APP2
[div, 0]

Status:
APP2: [2,1]
cons2: multiset
div: []
0: multiset

(14) Obligation:

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

APP0(cons0(n, x), y) → APP0(x, y)

The TRS R consists of the following rules:

div → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div → s0(div)
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(x, cons0(y, xs0)) → divL0(div, xs0)
shuffle0(nil0) → nil0
divL0(x, nil0) → x
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
less_leaves0(leaf0, cons0(w, z)) → true0
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

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

(15) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

APP0(cons0(n, x), y) → APP0(x, y)

Strictly oriented rules of the TRS R:

concat0(leaf0, y) → y
divL0(x, cons0(y, xs0)) → divL0(div, xs0)
shuffle0(nil0) → nil0
divL0(x, nil0) → x
less_leaves0(x, leaf0) → false0
less_leaves0(leaf0, cons0(w, z)) → true0
minus(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(APP0(x1, x2)) = x1 + x2   
POL(app0(x1, x2)) = x1 + x2   
POL(concat0(x1, x2)) = 2 + x1 + x2   
POL(cons0(x1, x2)) = 2 + x1 + x2   
POL(div) = 0   
POL(divL0(x1, x2)) = 1 + 2·x1 + x2   
POL(false0) = 0   
POL(leaf0) = 0   
POL(less_leaves0(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(minus(x1)) = 2 + 2·x1   
POL(nil0) = 0   
POL(reverse0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2 + x1   
POL(true0) = 0   
POL(xs0) = 0   

(16) Obligation:

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

div → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div → s0(div)
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
reverse0(nil0) → nil0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) YES

(19) Obligation:

Relative ADP Problem with
absolute ADPs:

div(s(x), s(y)) → s(DIV(minus(x, y), s(y)))

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(20) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:true = less_leaves_2 = app_2 = divL_2 = concat_2 = leaf = xs = shuffle_1 = nil = false = s_1 = reverse_1 = div_2 = 1 0 = minus_2 = 1 cons_2 = DIV_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
DIV(x1, x2)  =  DIV(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0
div(x1, x2)  =  div(x1)

Recursive path order with status [RPO].
Quasi-Precedence:

0 > [DIV2, s1, div1]

Status:
DIV2: multiset
s1: multiset
0: multiset
div1: multiset

(21) Obligation:

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

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

The TRS R consists of the following rules:

div(00) → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div(s0(x)) → s0(div(minus(x)))
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(x, cons0(y, xs0)) → divL0(div(x), xs0)
shuffle0(nil0) → nil0
divL0(x, nil0) → x
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
less_leaves0(leaf0, cons0(w, z)) → true0
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

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

(22) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

divL0(x, cons0(y, xs0)) → divL0(div(x), xs0)
divL0(x, nil0) → x
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
less_leaves0(leaf0, cons0(w, z)) → true0
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(DIV0(x1, x2)) = x1 + x2   
POL(app0(x1, x2)) = x1 + x2   
POL(concat0(x1, x2)) = x1 + x2   
POL(cons0(x1, x2)) = 2 + x1 + x2   
POL(div(x1)) = x1   
POL(divL0(x1, x2)) = 2 + x1 + x2   
POL(false0) = 0   
POL(leaf0) = 0   
POL(less_leaves0(x1, x2)) = 1 + 2·x1 + x2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(reverse0(x1)) = 1 + x1   
POL(s0(x1)) = 2·x1   
POL(shuffle0(x1)) = 2·x1   
POL(true0) = 2   
POL(xs0) = 2   

(23) Obligation:

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

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

The TRS R consists of the following rules:

div(00) → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div(s0(x)) → s0(div(minus(x)))
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
shuffle0(nil0) → nil0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
minus(x) → x

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

(24) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

concat0(leaf0, y) → y

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(DIV0(x1, x2)) = x1 + x2   
POL(app0(x1, x2)) = x1 + x2   
POL(concat0(x1, x2)) = x1 + x2   
POL(cons0(x1, x2)) = x1 + x2   
POL(div(x1)) = x1   
POL(divL0(x1, x2)) = x1 + x2   
POL(leaf0) = 2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(reverse0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2·x1   
POL(xs0) = 0   

(25) Obligation:

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

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

The TRS R consists of the following rules:

div(00) → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div(s0(x)) → s0(div(minus(x)))
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
shuffle0(nil0) → nil0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
minus(x) → x

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

(26) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

Strictly oriented rules of the TRS R:

div(00) → 00
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
shuffle0(nil0) → nil0
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2   
POL(DIV0(x1, x2)) = x1 + x2   
POL(app0(x1, x2)) = x1 + x2   
POL(concat0(x1, x2)) = 2·x1 + x2   
POL(cons0(x1, x2)) = 2 + x1 + x2   
POL(div(x1)) = 2 + x1   
POL(divL0(x1, x2)) = x1 + x2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(reverse0(x1)) = x1   
POL(s0(x1)) = 1 + x1   
POL(shuffle0(x1)) = 2 + 2·x1   
POL(xs0) = 0   

(27) Obligation:

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

divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div(s0(x)) → s0(div(minus(x)))
app0(cons0(n, x), y) → cons0(n, app0(x, y))
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(x) → x

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) YES

(30) Obligation:

Relative ADP Problem with
absolute ADPs:

concat(cons(u, v), y) → cons(u, CONCAT(v, y))

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(31) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:true = CONCAT_2 = less_leaves_2 = app_2 = concat_2 = divL_2 = leaf = xs = shuffle_1 = nil = false = s_1 = reverse_1 = div_2 = 0, 1 0 = minus_2 = 1 cons_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
CONCAT(x1, x2)  =  CONCAT(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
div(x1, x2)  =  div
s(x1)  =  x1
minus(x1, x2)  =  x1
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

cons2 > CONCAT2
[div, 0]

Status:
CONCAT2: [2,1]
cons2: multiset
div: multiset
0: multiset

(32) Obligation:

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

CONCAT0(cons0(u, v), y) → CONCAT0(v, y)

The TRS R consists of the following rules:

div → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div → s0(div)
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(x, cons0(y, xs0)) → divL0(div, xs0)
shuffle0(nil0) → nil0
divL0(x, nil0) → x
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
less_leaves0(leaf0, cons0(w, z)) → true0
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

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

(33) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

CONCAT0(cons0(u, v), y) → CONCAT0(v, y)

Strictly oriented rules of the TRS R:

concat0(leaf0, y) → y
divL0(x, cons0(y, xs0)) → divL0(div, xs0)
shuffle0(nil0) → nil0
divL0(x, nil0) → x
less_leaves0(x, leaf0) → false0
less_leaves0(leaf0, cons0(w, z)) → true0
minus(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(CONCAT0(x1, x2)) = x1 + x2   
POL(app0(x1, x2)) = x1 + x2   
POL(concat0(x1, x2)) = 2 + x1 + x2   
POL(cons0(x1, x2)) = 2 + x1 + x2   
POL(div) = 0   
POL(divL0(x1, x2)) = 1 + 2·x1 + x2   
POL(false0) = 0   
POL(leaf0) = 0   
POL(less_leaves0(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(minus(x1)) = 2 + 2·x1   
POL(nil0) = 0   
POL(reverse0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2 + x1   
POL(true0) = 0   
POL(xs0) = 0   

(34) Obligation:

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

div → 00
divL0(z, cons0(x, cons0(y, xs0))) → divL0(z, cons0(y, cons0(x, xs0)))
shuffle0(cons0(n, cons0(m, x))) → shuffle0(cons0(m, cons0(n, x)))
div → s0(div)
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
reverse0(nil0) → nil0
app0(cons0(n, x), y) → cons0(n, app0(x, y))
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
reverse0(cons0(n, x)) → app0(reverse0(x), cons0(n, nil0))
app0(nil0, y) → y
minus(s0(x)) → minus(x)
less_leaves0(cons0(u, v), cons0(w, z)) → less_leaves0(concat0(u, v), concat0(w, z))

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) YES

(37) Obligation:

Relative ADP Problem with
absolute ADPs:

less_leaves(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(38) RelADPDerelatifyingProof (EQUIVALENT transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

(39) Obligation:

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

LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))

The TRS R consists of the following rules:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( div(x1, x2) ) = x2 + 1

POL( 0 ) = 0

POL( s(x1) ) = x1

POL( divL(x1, x2) ) = x1 + 2x2

POL( cons(x1, x2) ) = x1 + x2 + 1

POL( xs ) = 1

POL( shuffle(x1) ) = x1 + 2

POL( minus(x1, x2) ) = x1 + x2

POL( concat(x1, x2) ) = x1 + x2

POL( leaf ) = 0

POL( nil ) = 0

POL( reverse(x1) ) = x1

POL( less_leaves(x1, x2) ) = 0

POL( false ) = 0

POL( app(x1, x2) ) = x1 + x2

POL( true ) = 0

POL( LESS_LEAVES(x1, x2) ) = x1 + x2


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(41) Obligation:

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

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) YES

(44) Obligation:

Relative ADP Problem with
absolute ADPs:

reverse(cons(n, x)) → app(REVERSE(x), cons(n, nil))

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(45) RelADPDerelatifyingProof (EQUIVALENT transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

(46) Obligation:

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

REVERSE(cons(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


REVERSE(cons(n, x)) → REVERSE(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( div(x1, x2) ) = x1 + 2

POL( 0 ) = 0

POL( s(x1) ) = x1 + 2

POL( divL(x1, x2) ) = x1 + x2 + 1

POL( cons(x1, x2) ) = x2 + 2

POL( xs ) = 1

POL( shuffle(x1) ) = 2x1 + 1

POL( minus(x1, x2) ) = x1

POL( concat(x1, x2) ) = 2x1 + x2 + 2

POL( leaf ) = 0

POL( nil ) = 0

POL( reverse(x1) ) = x1 + 1

POL( less_leaves(x1, x2) ) = 2

POL( false ) = 0

POL( app(x1, x2) ) = x1 + x2

POL( true ) = 0

POL( REVERSE(x1) ) = 2x1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(48) Obligation:

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

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → shuffle(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

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

(49) PisEmptyProof (EQUIVALENT transformation)

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

(50) YES

(51) Obligation:

Relative ADP Problem with
absolute ADPs:

shuffle(cons(n, x)) → cons(n, SHUFFLE(reverse(x)))

and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → SHUFFLE(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(52) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


shuffle(cons(n, x)) → cons(n, SHUFFLE(reverse(x)))

Relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(APP(x1, x2)) = 2·x1·x2 + 2·x2   
POL(CONCAT(x1, x2)) = 2 + 2·x1   
POL(DIV(x1, x2)) = 2 + x2   
POL(DIVL(x1, x2)) = 2 + 3·x1 + x1·x2   
POL(LESS_LEAVES(x1, x2)) = 2   
POL(MINUS(x1, x2)) = 2   
POL(REVERSE(x1)) = 2 + 2·x1   
POL(SHUFFLE(x1)) = 2 + 2·x1   
POL(app(x1, x2)) = x1 + x2   
POL(concat(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(div(x1, x2)) = 0   
POL(divL(x1, x2)) = 2 + x1 + 2·x2   
POL(false) = 0   
POL(leaf) = 0   
POL(less_leaves(x1, x2)) = 3·x2   
POL(minus(x1, x2)) = 2·x1   
POL(nil) = 0   
POL(reverse(x1)) = x1   
POL(s(x1)) = 3·x1   
POL(shuffle(x1)) = 3·x1   
POL(true) = 1   
POL(xs) = 3   

(53) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

div(0, s(y)) → 0
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
shuffle(cons(n, cons(m, x))) → SHUFFLE(cons(m, cons(n, x)))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
divL(x, nil) → x
reverse(nil) → nil
less_leaves(x, leaf) → false
app(cons(n, x), y) → cons(n, app(x, y))
shuffle(cons(n, x)) → cons(n, shuffle(reverse(x)))
less_leaves(leaf, cons(w, z)) → true
reverse(cons(n, x)) → app(reverse(x), cons(n, nil))
app(nil, y) → y
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
less_leaves(cons(u, v), cons(w, z)) → less_leaves(concat(u, v), concat(w, z))

(54) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(55) YES