YES Termination proof of AG_#3.53_rand.trs

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 (⇒, 241 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 6 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 12 ms)
        ↳15 QDP
        ↳16 PisEmptyProof (⇔, 0 ms)
        ↳17 YES
    ↳18 RelADPP
        ↳19 RelADPCleverAfsProof (⇒, 230 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 MRRProof (⇔, 8 ms)
        ↳24 QDP
        ↳25 MRRProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 MRRProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 PisEmptyProof (⇔, 0 ms)
        ↳30 YES
    ↳31 RelADPP
        ↳32 RelADPDerelatifyingProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 QDPOrderProof (⇔, 46 ms)
        ↳35 QDP
        ↳36 PisEmptyProof (⇔, 0 ms)
        ↳37 YES
    ↳38 RelADPP
        ↳39 RelADPDerelatifyingProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 QDPOrderProof (⇔, 28 ms)
        ↳42 QDP
        ↳43 PisEmptyProof (⇔, 0 ms)
        ↳44 YES
    ↳45 RelADPP
        ↳46 RelADPDerelatifyingProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 QDPOrderProof (⇔, 26 ms)
        ↳49 QDP
        ↳50 PisEmptyProof (⇔, 0 ms)
        ↳51 YES
    ↳52 RelADPP
        ↳53 RelADPDerelatifyingProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 QDPOrderProof (⇔, 39 ms)
        ↳56 QDP
        ↳57 PisEmptyProof (⇔, 0 ms)
        ↳58 YES
    ↳59 RelADPP
        ↳60 RelADPCleverAfsProof (⇒, 237 ms)
        ↳61 QDP
        ↳62 MRRProof (⇔, 0 ms)
        ↳63 QDP
        ↳64 MRRProof (⇔, 8 ms)
        ↳65 QDP
        ↳66 MRRProof (⇔, 0 ms)
        ↳67 QDP
        ↳68 MRRProof (⇔, 0 ms)
        ↳69 QDP
        ↳70 MRRProof (⇔, 0 ms)
        ↳71 QDP
        ↳72 QDPOrderProof (⇔, 16 ms)
        ↳73 QDP
        ↳74 PisEmptyProof (⇔, 0 ms)
        ↳75 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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
reverse(nil) → nil
reverse(add(n, x)) → app(reverse(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(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:

rand(x) → rand(s(x))
rand(x) → 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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))
quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))
app(nil, y) → y
app(add(n, x), y) → add(n, APP(x, y))
reverse(nil) → nil
reverse(add(n, x)) → APP(reverse(x), add(n, nil))
reverse(add(n, x)) → app(REVERSE(x), add(n, nil))
shuffle(nil) → nil
shuffle(add(n, x)) → add(n, SHUFFLE(reverse(x)))
shuffle(add(n, x)) → add(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:

rand(x) → RAND(s(x))
rand(x) → 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:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(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 = 1 true = less_leaves_2 = add_2 = 0 app_2 = 0 concat_2 = 0 leaf = shuffle_1 = 0 rand_1 = quot_2 = nil = false = s_1 = reverse_1 = 0 0 = minus_2 = 1 cons_2 = 0
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(x1)
s(x1)  =  s(x1)
quot(x1, x2)  =  quot(x1, x2)
minus(x1, x2)  =  minus(x1)
concat(x1, x2)  =  x2
leaf  =  leaf
cons(x1, x2)  =  x2
shuffle(x1)  =  shuffle
nil  =  nil
add(x1, x2)  =  x2
reverse(x1)  =  reverse
0  =  0
less_leaves(x1, x2)  =  less_leaves(x1, x2)
false  =  false
app(x1, x2)  =  x2
true  =  true

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

quot2 > s1 > MINUS1
quot2 > s1 > minus1
quot2 > 0
[shuffle, nil, reverse]
lessleaves2 > false
lessleaves2 > true

Status:
MINUS1: multiset
s1: [1]
quot2: [1,2]
minus1: multiset
leaf: multiset
shuffle: []
nil: multiset
reverse: []
0: multiset
lessleaves2: [2,1]
false: multiset
true: multiset

(7) Obligation:

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

MINUS(s0(x)) → MINUS(x)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle → nil0
shuffle → add(shuffle)
quot0(00, s0(y)) → 00
reverse → nil0
less_leaves0(x, leaf0) → false0
app(y) → add(app(y))
less_leaves0(leaf0, cons(z)) → true0
app(y) → y
reverse → app(add(nil0))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons(v), cons(z)) → less_leaves0(concat(v), concat(z))
rand0(x) → x

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:

shuffle → nil0
reverse → nil0
reverse → app(add(nil0))
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = 2·x1   
POL(add(x1)) = x1   
POL(app(x1)) = x1   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(false0) = 0   
POL(leaf0) = 0   
POL(less_leaves0(x1, x2)) = x1 + x2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot0(x1, x2)) = x1 + x2   
POL(rand0(x1)) = 2 + x1   
POL(reverse) = 2   
POL(s0(x1)) = x1   
POL(shuffle) = 2   
POL(true0) = 0   

(9) Obligation:

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

MINUS(s0(x)) → MINUS(x)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
quot0(00, s0(y)) → 00
less_leaves0(x, leaf0) → false0
app(y) → add(app(y))
less_leaves0(leaf0, cons(z)) → true0
app(y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons(v), cons(z)) → less_leaves0(concat(v), concat(z))

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

(10) 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:

app(y) → y

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = x1   
POL(add(x1)) = x1   
POL(app(x1)) = 2 + x1   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(false0) = 0   
POL(leaf0) = 0   
POL(less_leaves0(x1, x2)) = x1 + x2   
POL(minus(x1)) = x1   
POL(quot0(x1, x2)) = x1 + x2   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle) = 0   
POL(true0) = 0   

(11) Obligation:

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

MINUS(s0(x)) → MINUS(x)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
quot0(00, s0(y)) → 00
less_leaves0(x, leaf0) → false0
app(y) → add(app(y))
less_leaves0(leaf0, cons(z)) → true0
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons(v), cons(z)) → less_leaves0(concat(v), concat(z))

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

(12) 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:

quot0(00, s0(y)) → 00
less_leaves0(x, leaf0) → false0
less_leaves0(leaf0, cons(z)) → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = 2·x1   
POL(add(x1)) = x1   
POL(app(x1)) = x1   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(false0) = 1   
POL(leaf0) = 2   
POL(less_leaves0(x1, x2)) = 2·x1 + 2·x2   
POL(minus(x1)) = x1   
POL(quot0(x1, x2)) = 1 + x1 + x2   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle) = 0   
POL(true0) = 2   

(13) Obligation:

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

MINUS(s0(x)) → MINUS(x)

The TRS R consists of the following rules:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
app(y) → add(app(y))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons(v), cons(z)) → less_leaves0(concat(v), concat(z))

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

(14) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


MINUS(s0(x)) → MINUS(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  x1
s0(x1)  =  s0(x1)
quot0(x1, x2)  =  quot0(x1)
minus(x1)  =  minus(x1)
concat(x1)  =  x1
cons(x1)  =  x1
rand0(x1)  =  rand0
shuffle  =  shuffle
add(x1)  =  x1
app(x1)  =  app(x1)
less_leaves0(x1, x2)  =  less_leaves0(x2)

Recursive path order with status [RPO].
Quasi-Precedence:
quot01 > s01 > minus1
rand0 > s01 > minus1

Status:
s01: [1]
quot01: [1]
minus1: multiset
rand0: []
shuffle: multiset
app1: multiset
lessleaves01: [1]


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

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
app(y) → add(app(y))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons(v), cons(z)) → less_leaves0(concat(v), concat(z))

(15) Obligation:

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

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
app(y) → add(app(y))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves0(cons(v), cons(z)) → less_leaves0(concat(v), concat(z))

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) YES

(18) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(19) 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 = 0 add_2 = 0 app_2 = 0 concat_2 = leaf = shuffle_1 = 0 rand_1 = quot_2 = 1 nil = false = s_1 = reverse_1 = 0 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)
quot(x1, x2)  =  quot(x1)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
concat(x1, x2)  =  concat(x1, x2)
leaf  =  leaf
shuffle(x1)  =  shuffle
nil  =  nil
add(x1, x2)  =  x2
reverse(x1)  =  reverse
0  =  0
less_leaves(x1, x2)  =  x2
false  =  false
app(x1, x2)  =  x2
true  =  true

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

CONCAT2 > false
[cons2, concat2, leaf, true] > false
quot1 > s1 > minus1 > false
quot1 > 0 > false
shuffle > reverse > nil > false

Status:
CONCAT2: multiset
cons2: [1,2]
quot1: multiset
s1: [1]
minus1: [1]
concat2: [1,2]
leaf: multiset
shuffle: []
nil: multiset
reverse: []
0: multiset
false: multiset
true: multiset

(20) 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:

quot(s0(x)) → s0(quot(minus(x)))
concat0(leaf0, y) → y
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
rand0(x) → rand0(s0(x))
shuffle → nil0
shuffle → add(shuffle)
quot(00) → 00
reverse → nil0
less_leaves(leaf0) → false0
app(y) → add(app(y))
less_leaves(cons0(w, z)) → true0
app(y) → y
reverse → app(add(nil0))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons0(w, z)) → less_leaves(concat0(w, z))
rand0(x) → x

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

(21) 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
reverse → nil0
less_leaves(leaf0) → false0
reverse → app(add(nil0))
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(CONCAT0(x1, x2)) = x1 + x2   
POL(add(x1)) = x1   
POL(app(x1)) = x1   
POL(concat0(x1, x2)) = 2·x1 + x2   
POL(cons0(x1, x2)) = 2·x1 + x2   
POL(false0) = 0   
POL(leaf0) = 1   
POL(less_leaves(x1)) = x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(reverse) = 2   
POL(s0(x1)) = x1   
POL(shuffle) = 0   
POL(true0) = 0   

(22) 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:

quot(s0(x)) → s0(quot(minus(x)))
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
rand0(x) → rand0(s0(x))
shuffle → nil0
shuffle → add(shuffle)
quot(00) → 00
app(y) → add(app(y))
less_leaves(cons0(w, z)) → true0
app(y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons0(w, z)) → less_leaves(concat0(w, z))

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

(23) 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:

app(y) → y

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(CONCAT0(x1, x2)) = 2·x1 + x2   
POL(add(x1)) = x1   
POL(app(x1)) = 2 + x1   
POL(concat0(x1, x2)) = 2·x1 + x2   
POL(cons0(x1, x2)) = 2·x1 + x2   
POL(less_leaves(x1)) = 1 + x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle) = 0   
POL(true0) = 1   

(24) 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:

quot(s0(x)) → s0(quot(minus(x)))
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
rand0(x) → rand0(s0(x))
shuffle → nil0
shuffle → add(shuffle)
quot(00) → 00
app(y) → add(app(y))
less_leaves(cons0(w, z)) → true0
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons0(w, z)) → less_leaves(concat0(w, z))

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

(25) 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:

shuffle → nil0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(CONCAT0(x1, x2)) = x1 + x2   
POL(add(x1)) = x1   
POL(app(x1)) = x1   
POL(concat0(x1, x2)) = 2·x1 + x2   
POL(cons0(x1, x2)) = 2·x1 + x2   
POL(less_leaves(x1)) = 1 + x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle) = 1   
POL(true0) = 1   

(26) 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:

quot(s0(x)) → s0(quot(minus(x)))
concat0(cons0(u, v), y) → cons0(u, concat0(v, y))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
quot(00) → 00
app(y) → add(app(y))
less_leaves(cons0(w, z)) → true0
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons0(w, z)) → less_leaves(concat0(w, z))

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

(27) 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(cons0(u, v), y) → cons0(u, concat0(v, y))
quot(00) → 00
less_leaves(cons0(w, z)) → true0
less_leaves(cons0(w, z)) → less_leaves(concat0(w, z))

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(CONCAT0(x1, x2)) = 2·x1 + x2   
POL(add(x1)) = x1   
POL(app(x1)) = x1   
POL(concat0(x1, x2)) = 1 + 2·x1 + x2   
POL(cons0(x1, x2)) = 2 + 2·x1 + x2   
POL(less_leaves(x1)) = 2 + x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = 2·x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle) = 0   
POL(true0) = 2   

(28) Obligation:

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

quot(s0(x)) → s0(quot(minus(x)))
rand0(x) → rand0(s0(x))
shuffle → add(shuffle)
app(y) → add(app(y))
minus(s0(x)) → minus(x)
minus(x) → x

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

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) YES

(31) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(32) 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.

(33) 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:

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

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

(34) 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( quot(x1, x2) ) = x1 + x2 + 1

POL( s(x1) ) = x1

POL( minus(x1, x2) ) = x1

POL( concat(x1, x2) ) = 2x1 + x2 + 1

POL( leaf ) = 0

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

POL( rand(x1) ) = 2x1

POL( shuffle(x1) ) = 2x1

POL( nil ) = 1

POL( add(x1, x2) ) = 2

POL( reverse(x1) ) = 2x1 + 2

POL( 0 ) = 1

POL( less_leaves(x1, x2) ) = 2x1 + x2 + 1

POL( false ) = 0

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

POL( true ) = 2

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


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

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

(35) Obligation:

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

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

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

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) YES

(38) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(39) 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.

(40) Obligation:

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

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

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

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

(41) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL( s(x1) ) = x1

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

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

POL( leaf ) = 0

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

POL( rand(x1) ) = x1

POL( shuffle(x1) ) = x1

POL( nil ) = 0

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

POL( reverse(x1) ) = x1

POL( 0 ) = 1

POL( less_leaves(x1, x2) ) = x1 + 2x2 + 1

POL( false ) = 0

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

POL( true ) = 0

POL( APP(x1, x2) ) = x1


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

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

(42) Obligation:

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

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

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

(43) PisEmptyProof (EQUIVALENT transformation)

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

(44) YES

(45) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(46) 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.

(47) Obligation:

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

REVERSE(add(n, x)) → REVERSE(x)

The TRS R consists of the following rules:

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

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

(48) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL( s(x1) ) = x1

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

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

POL( leaf ) = 0

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

POL( rand(x1) ) = x1

POL( shuffle(x1) ) = x1

POL( nil ) = 0

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

POL( reverse(x1) ) = x1

POL( 0 ) = 1

POL( less_leaves(x1, x2) ) = x1 + 2x2 + 1

POL( false ) = 0

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

POL( true ) = 0

POL( REVERSE(x1) ) = x1


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

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

(49) Obligation:

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

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

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

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) YES

(52) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(53) 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.

(54) Obligation:

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

SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))

The TRS R consists of the following rules:

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

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

(55) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL( s(x1) ) = x1

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

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

POL( leaf ) = 0

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

POL( rand(x1) ) = 2x1

POL( shuffle(x1) ) = x1 + 1

POL( nil ) = 0

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

POL( reverse(x1) ) = x1

POL( 0 ) = 0

POL( less_leaves(x1, x2) ) = 2x1 + x2 + 1

POL( false ) = 0

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

POL( true ) = 0

POL( SHUFFLE(x1) ) = x1 + 2


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

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

(56) Obligation:

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

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

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

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) YES

(59) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

rand(x) → rand(s(x))
rand(x) → x

(60) 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 = QUOT_2 = less_leaves_2 = 0 add_2 = 0, 1 app_2 = 0 concat_2 = 0 leaf = shuffle_1 = rand_1 = quot_2 = 1 nil = false = s_1 = reverse_1 = 0 0 = minus_2 = 1 cons_2 = 0
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
quot(x1, x2)  =  quot(x1)
concat(x1, x2)  =  x2
leaf  =  leaf
cons(x1, x2)  =  x2
shuffle(x1)  =  shuffle(x1)
nil  =  nil
add(x1, x2)  =  add
reverse(x1)  =  reverse
0  =  0
less_leaves(x1, x2)  =  x2
false  =  false
app(x1, x2)  =  x2
true  =  true

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

QUOT2 > s1 > [nil, add, true]
quot1 > s1 > [nil, add, true]
[leaf, false] > [nil, add, true]
shuffle1 > [nil, add, true]
reverse > [nil, add, true]
0 > [nil, add, true]

Status:
QUOT2: [2,1]
s1: multiset
quot1: multiset
leaf: multiset
shuffle1: [1]
nil: multiset
add: multiset
reverse: []
0: multiset
false: multiset
true: multiset

(61) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle0(nil0) → nil0
shuffle0(add) → add
quot(00) → 00
reverse → nil0
less_leaves(leaf0) → false0
app(y) → add
less_leaves(cons(z)) → true0
app(y) → y
reverse → app(add)
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))
rand0(x) → x

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

(62) 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:

reverse → nil0
less_leaves(leaf0) → false0
reverse → app(add)
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = x1 + x2   
POL(add) = 0   
POL(app(x1)) = 2·x1   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(false0) = 0   
POL(leaf0) = 1   
POL(less_leaves(x1)) = x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(reverse) = 2   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = x1   
POL(true0) = 0   

(63) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle0(nil0) → nil0
shuffle0(add) → add
quot(00) → 00
app(y) → add
less_leaves(cons(z)) → true0
app(y) → y
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

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

(64) 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:

app(y) → add
app(y) → y

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = x1 + x2   
POL(add) = 0   
POL(app(x1)) = 2 + x1   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(less_leaves(x1)) = 1 + x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2·x1   
POL(true0) = 1   

(65) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle0(nil0) → nil0
shuffle0(add) → add
quot(00) → 00
less_leaves(cons(z)) → true0
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

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

(66) 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:

less_leaves(cons(z)) → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = x1 + x2   
POL(add) = 0   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(less_leaves(x1)) = 2 + x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2·x1   
POL(true0) = 1   

(67) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle0(nil0) → nil0
shuffle0(add) → add
quot(00) → 00
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

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

(68) 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:

shuffle0(add) → add
quot(00) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(QUOT0(x1, x2)) = x1 + x2   
POL(add) = 1   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(less_leaves(x1)) = x1   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(quot(x1)) = 2·x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2·x1   

(69) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
shuffle0(nil0) → nil0
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

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

(70) 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:

shuffle0(nil0) → nil0

Used ordering: Polynomial interpretation [POLO]:

POL(QUOT0(x1, x2)) = x1 + x2   
POL(concat(x1)) = x1   
POL(cons(x1)) = x1   
POL(less_leaves(x1)) = 2·x1   
POL(minus(x1)) = x1   
POL(nil0) = 2   
POL(quot(x1)) = 2·x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(shuffle0(x1)) = 2·x1   

(71) Obligation:

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

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

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

(72) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( quot(x1) ) = 2x1 + 1

POL( s0(x1) ) = 2x1 + 1

POL( minus(x1) ) = x1

POL( concat(x1) ) = x1

POL( cons(x1) ) = x1

POL( rand0(x1) ) = max{0, -2}

POL( less_leaves(x1) ) = x1 + 2

POL( QUOT0(x1, x2) ) = 2x1 + 2x2 + 2


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

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

(73) Obligation:

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

quot(s0(x)) → s0(quot(minus(x)))
concat(y) → y
concat(y) → cons(concat(y))
rand0(x) → rand0(s0(x))
minus(s0(x)) → minus(x)
minus(x) → x
less_leaves(cons(z)) → less_leaves(concat(z))

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

(74) PisEmptyProof (EQUIVALENT transformation)

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

(75) YES