Termination proof of AProVE_24_combination

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 (⇒, 43 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 RelADPP

        ↳15 RelADPCleverAfsProof (⇒, 30 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔, 0 ms)

        ↳20 YES

    ↳21 RelADPP

        ↳22 RelADPCleverAfsProof (⇒, 51 ms)

        ↳23 QDP

        ↳24 MRRProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 MRRProof (⇔, 0 ms)

        ↳27 QDP

        ↳28 PisEmptyProof (⇔, 0 ms)

        ↳29 YES

    ↳30 RelADPP

        ↳31 RelADPCleverAfsProof (⇒, 72 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 (⇔, 23 ms)

        ↳41 QDP

        ↳42 PisEmptyProof (⇔, 0 ms)

        ↳43 YES

    ↳44 RelADPP

        ↳45 RelADPDerelatifyingProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 QDPOrderProof (⇔, 26 ms)

        ↳48 QDP

        ↳49 PisEmptyProof (⇔, 0 ms)

        ↳50 YES

    ↳51 RelADPP

        ↳52 RelADPReductionPairProof (⇔, 60 ms)

        ↳53 RelADPP

        ↳54 DAbsisEmptyProof (⇔, 0 ms)

        ↳55 YES

    ↳56 RelADPP

        ↳57 RelADPReductionPairProof (⇔, 69 ms)

        ↳58 RelADPP

        ↳59 DAbsisEmptyProof (⇔, 0 ms)

        ↳60 YES

    ↳61 RelADPP

        ↳62 RelADPReductionPairProof (⇔, 56 ms)

        ↳63 RelADPP

        ↳64 DAbsisEmptyProof (⇔, 0 ms)

        ↳65 YES

    ↳66 RelADPP

        ↳67 RelADPReductionPairProof (⇔, 67 ms)

        ↳68 RelADPP

        ↳69 DAbsisEmptyProof (⇔, 0 ms)

        ↳70 YES

    ↳71 RelADPP

        ↳72 RelADPReductionPairProof (⇔, 61 ms)

        ↳73 RelADPP

        ↳74 DAbsisEmptyProof (⇔, 0 ms)

        ↳75 YES

    ↳76 RelADPP

        ↳77 RelADPReductionPairProof (⇔, 69 ms)

        ↳78 RelADPP

        ↳79 DAbsisEmptyProof (⇔, 0 ms)

        ↳80 YES

    ↳81 RelADPP

        ↳82 RelADPReductionPairProof (⇔, 70 ms)

        ↳83 RelADPP

        ↳84 DAbsisEmptyProof (⇔, 0 ms)

        ↳85 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:

shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → cons(x, xs)
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))

(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:

shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
consSwap(x, xs) → cons(x, xs)
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, cons(y, xs)) → cons(y, CONSSWAP(x, xs))

(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,
6 Lassos,
Result: This relative DT problem is equivalent to 13 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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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 = concat_2 = leaf = divL_2 = xs = shuffle_1 = nil = false = s_1 = reverse_1 = div_2 = 0 = minus_2 = 1 cons_2 = consSwap_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
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
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(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
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 rules of the TRS R:

divL0(x, cons0(y, xs0)) → divL0(div0(x, y), 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(MINUS(x₁)) = 2·x₁
POL(app0(x₁, x₂)) = x₁ + x₂
POL(concat0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + x₁ + x₂
POL(div0(x₁, x₂)) = x₁ + x₂
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)) = x₁
POL(shuffle0(x₁)) = 2·x₁
POL(true0) = 2
POL(xs0) = 0

(9) 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
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
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(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
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.

(10) MRRProof (EQUIVALENT transformation)

MINUS(s0(y)) → MINUS(y)

Strictly oriented rules of the TRS R:

div0(00, s0(y)) → 00
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
concat0(leaf0, y) → y
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(MINUS(x₁)) = 2·x₁
POL(app0(x₁, x₂)) = x₁ + x₂
POL(concat0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2·x₁ + x₂
POL(div0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(leaf0) = 2
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(reverse0(x₁)) = x₁
POL(s0(x₁)) = 2 + x₁
POL(shuffle0(x₁)) = 2·x₁
POL(xs0) = 2

(11) Obligation:

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

shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
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(x) → x

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) YES

(14) 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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(15) 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 = concat_2 = leaf = divL_2 = xs = shuffle_1 = APP_2 = nil = false = s_1 = reverse_1 = div_2 = 0, 1 0 = minus_2 = 1 cons_2 = consSwap_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

(16) 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
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div → s0(div)
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
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.

(17) MRRProof (EQUIVALENT transformation)

APP0(cons0(n, x), y) → APP0(x, y)

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(APP0(x₁, x₂)) = x₁ + x₂
POL(app0(x₁, x₂)) = x₁ + x₂
POL(concat0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + x₁ + x₂
POL(div) = 2
POL(divL0(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(false0) = 0
POL(leaf0) = 0
POL(less_leaves0(x₁, x₂)) = 2·x₁ + x₂
POL(minus(x₁)) = 1 + x₁
POL(nil0) = 0
POL(reverse0(x₁)) = x₁
POL(s0(x₁)) = x₁
POL(shuffle0(x₁)) = 2·x₁
POL(true0) = 0
POL(xs0) = 2

(18) Obligation:

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

div → 00
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div → s0(div)
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
shuffle0(nil0) → nil0
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
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(s0(x)) → minus(x)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES

(21) 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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(22) 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 = concat_2 = leaf = divL_2 = xs = shuffle_1 = nil = false = s_1 = reverse_1 = div_2 = 1 0 = minus_2 = 1 cons_2 = consSwap_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

(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
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div(s0(x)) → s0(div(minus(x)))
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
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.

(24) MRRProof (EQUIVALENT transformation)

div(00) → 00
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(x, nil0) → x
shuffle0(cons0(n, x)) → cons0(n, shuffle0(reverse0(x)))
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)) = 2·x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(div(x₁)) = 2 + x₁
POL(divL0(x₁, x₂)) = 2 + x₁ + x₂
POL(false0) = 0
POL(leaf0) = 0
POL(less_leaves0(x₁, x₂)) = 2·x₁ + x₂
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(reverse0(x₁)) = x₁
POL(s0(x₁)) = x₁
POL(shuffle0(x₁)) = 2·x₁
POL(true0) = 0
POL(xs0) = 2

(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:

shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div(s0(x)) → s0(div(minus(x)))
concat0(leaf0, y) → y
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
divL0(x, cons0(y, xs0)) → divL0(div(x), xs0)
shuffle0(nil0) → nil0
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
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(s0(x)) → minus(x)
minus(x) → x

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

(26) MRRProof (EQUIVALENT transformation)

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

Strictly oriented rules of the TRS R:

concat0(leaf0, y) → y
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
divL0(x, cons0(y, xs0)) → divL0(div(x), xs0)
shuffle0(nil0) → nil0
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(DIV0(x₁, x₂)) = x₁ + x₂
POL(app0(x₁, x₂)) = x₁ + 2·x₂
POL(concat0(x₁, x₂)) = 2 + x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2·x₁ + 2·x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = x₁ + x₂
POL(false0) = 0
POL(leaf0) = 0
POL(less_leaves0(x₁, x₂)) = x₁ + x₂
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(reverse0(x₁)) = 2·x₁
POL(s0(x₁)) = 2 + 2·x₁
POL(shuffle0(x₁)) = 1 + 2·x₁
POL(xs0) = 2

(27) Obligation:

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

shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div(s0(x)) → s0(div(minus(x)))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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 = leaf = divL_2 = xs = shuffle_1 = nil = false = s_1 = reverse_1 = div_2 = 0, 1 0 = minus_2 = 1 cons_2 = consSwap_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
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div → s0(div)
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
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)

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

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(CONCAT0(x₁, x₂)) = x₁ + x₂
POL(app0(x₁, x₂)) = x₁ + x₂
POL(concat0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + x₁ + x₂
POL(div) = 2
POL(divL0(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(false0) = 0
POL(leaf0) = 0
POL(less_leaves0(x₁, x₂)) = 2·x₁ + x₂
POL(minus(x₁)) = 1 + x₁
POL(nil0) = 0
POL(reverse0(x₁)) = x₁
POL(s0(x₁)) = x₁
POL(shuffle0(x₁)) = 2·x₁
POL(true0) = 0
POL(xs0) = 2

(34) Obligation:

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

div → 00
shuffle0(cons0(x, xs0)) → shuffle0(consSwap0(x, xs0))
consSwap0(x, xs0) → cons0(x, xs0)
div → s0(div)
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
divL0(z, cons0(x, xs0)) → divL0(z, consSwap0(x, xs0))
consSwap0(x, cons0(y, xs0)) → cons0(y, consSwap0(x, xs0))
shuffle0(nil0) → nil0
reverse0(nil0) → nil0
less_leaves0(x, leaf0) → false0
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(s0(x)) → minus(x)

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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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) ) = max{0, -2}

POL( 0 ) = 0

POL( s(x₁) ) = 2x₁

POL( shuffle(x₁) ) = 2x₁ + 1

POL( cons(x₁, x₂) ) = 2x₁ + x₂ + 2

POL( xs ) = 0

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

POL( minus(x₁, x₂) ) = 2x₁

POL( concat(x₁, x₂) ) = x₁ + x₂

POL( leaf ) = 0

POL( divL(x₁, x₂) ) = x₁ + 2

POL( nil ) = 0

POL( reverse(x₁) ) = x₁ + 1

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

POL( false ) = 0

POL( app(x₁, x₂) ) = x₁ + x₂

POL( true ) = 2

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

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

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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) ) = max{0, x₁ + x₂ - 2}

POL( 0 ) = 0

POL( s(x₁) ) = x₁ + 2

POL( shuffle(x₁) ) = 2x₁ + 2

POL( cons(x₁, x₂) ) = 2x₁ + x₂ + 2

POL( xs ) = 2

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

POL( minus(x₁, x₂) ) = x₁

POL( concat(x₁, x₂) ) = 2x₁ + x₂ + 2

POL( leaf ) = 0

POL( divL(x₁, x₂) ) = 2x₁ + x₂

POL( nil ) = 0

POL( reverse(x₁) ) = x₁ + 1

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

POL( false ) = 0

POL( app(x₁, x₂) ) = x₁ + x₂

POL( true ) = 2

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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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 + 2·x₁ + 2·x₁·x₂ + 2·x₂
POL(CONCAT(x₁, x₂)) = 2 + x₁ + x₁·x₂
POL(CONSSWAP(x₁, x₂)) = 0
POL(DIV(x₁, x₂)) = 2
POL(DIVL(x₁, x₂)) = 2 + 2·x₁
POL(LESS_LEAVES(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2
POL(REVERSE(x₁)) = 2 + 2·x₁
POL(SHUFFLE(x₁)) = 2 + 3·x₁
POL(app(x₁, x₂)) = x₁ + x₂
POL(concat(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(cons(x₁, x₂)) = 1 + x₂
POL(consSwap(x₁, x₂)) = 1 + x₂
POL(div(x₁, x₂)) = 0
POL(divL(x₁, x₂)) = 2 + 2·x₁
POL(false) = 0
POL(leaf) = 0
POL(less_leaves(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 2·x₁
POL(nil) = 0
POL(reverse(x₁)) = x₁
POL(s(x₁)) = x₁
POL(shuffle(x₁)) = 3 + x₁
POL(true) = 0
POL(xs) = 3

(53) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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

(56) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(57) 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
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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 + 2·x₁ + 2·x₁·x₂ + 2·x₂
POL(CONCAT(x₁, x₂)) = 2 + x₁ + x₁·x₂
POL(CONSSWAP(x₁, x₂)) = 0
POL(DIV(x₁, x₂)) = 2
POL(DIVL(x₁, x₂)) = 2 + 2·x₁
POL(LESS_LEAVES(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2
POL(REVERSE(x₁)) = 2 + 2·x₁
POL(SHUFFLE(x₁)) = 2 + 3·x₁
POL(app(x₁, x₂)) = x₁ + x₂
POL(concat(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(cons(x₁, x₂)) = 1 + x₂
POL(consSwap(x₁, x₂)) = 1 + x₂
POL(div(x₁, x₂)) = 0
POL(divL(x₁, x₂)) = 2 + 2·x₁
POL(false) = 0
POL(leaf) = 0
POL(less_leaves(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 2·x₁
POL(nil) = 0
POL(reverse(x₁)) = x₁
POL(s(x₁)) = x₁
POL(shuffle(x₁)) = 3 + x₁
POL(true) = 0
POL(xs) = 3

(58) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(59) DAbsisEmptyProof (EQUIVALENT transformation)

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

(60) YES

(61) Obligation:

Relative ADP Problem with
absolute ADPs:

shuffle(cons(n, x)) → cons(n, shuffle(REVERSE(x)))

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(62) 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
consSwap(x, xs) → 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(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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
POL(CONCAT(x₁, x₂)) = 2 + 2·x₂
POL(CONSSWAP(x₁, x₂)) = 0
POL(DIV(x₁, x₂)) = 2
POL(DIVL(x₁, x₂)) = 2
POL(LESS_LEAVES(x₁, x₂)) = 2 + 2·x₁·x₂
POL(MINUS(x₁, x₂)) = 0
POL(REVERSE(x₁)) = 0
POL(SHUFFLE(x₁)) = 1 + 3·x₁
POL(app(x₁, x₂)) = 3 + x₂
POL(concat(x₁, x₂)) = 3·x₁ + x₂
POL(cons(x₁, x₂)) = 3
POL(consSwap(x₁, x₂)) = 3 + x₂
POL(div(x₁, x₂)) = x₁
POL(divL(x₁, x₂)) = 2 + 2·x₁ + 3·x₂
POL(false) = 1
POL(leaf) = 0
POL(less_leaves(x₁, x₂)) = 2
POL(minus(x₁, x₂)) = x₁
POL(nil) = 1
POL(reverse(x₁)) = 2 + 2·x₁
POL(s(x₁)) = 3·x₁
POL(shuffle(x₁)) = 3
POL(true) = 0
POL(xs) = 0

(63) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(64) DAbsisEmptyProof (EQUIVALENT transformation)

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

(65) YES

(66) Obligation:

Relative ADP Problem with
absolute ADPs:

shuffle(nil) → nil

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
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))

(67) 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(nil) → nil

Relative ADPs:

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

(68) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
concat(leaf, y) → y
shuffle(cons(x, xs)) → SHUFFLE(CONSSWAP(x, xs))
concat(cons(u, v), y) → cons(u, concat(v, y))
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(69) DAbsisEmptyProof (EQUIVALENT transformation)

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

(70) YES

(71) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, cons(y, xs)) → DIVL(div(x, y), xs)

and relative ADPs:

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, xs) → 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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(72) 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:

divL(x, cons(y, xs)) → DIVL(div(x, y), xs)

Relative ADPs:

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

(73) Obligation:

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

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, xs) → 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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(74) DAbsisEmptyProof (EQUIVALENT transformation)

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

(75) YES

(76) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, nil) → x

and relative ADPs:

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, xs) → 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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, cons(y, xs)) → divL(div(x, y), xs)
shuffle(nil) → nil
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))

(77) 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:

divL(x, nil) → x

Relative ADPs:

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

(78) Obligation:

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

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, xs) → 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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(79) DAbsisEmptyProof (EQUIVALENT transformation)

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

(80) YES

(81) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, cons(y, xs)) → divL(DIV(x, y), xs)

and relative ADPs:

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, xs) → 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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(82) 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:

divL(x, cons(y, xs)) → divL(DIV(x, y), xs)

Relative ADPs:

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

(83) Obligation:

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

div(0, s(y)) → 0
shuffle(cons(x, xs)) → shuffle(consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, xs) → 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))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
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))

(84) DAbsisEmptyProof (EQUIVALENT transformation)

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