Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

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))

The TRS R 2 is

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 signature Sigma is {less_leaves, false, true}

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
APP(add(n, x), y) → APP(x, y)
REVERSE(add(n, x)) → APP(reverse(x), add(n, nil))
REVERSE(add(n, x)) → REVERSE(x)
SHUFFLE(add(n, x)) → SHUFFLE(reverse(x))
SHUFFLE(add(n, x)) → REVERSE(x)
CONCAT(cons(u, v), y) → CONCAT(v, y)
LESS_LEAVES(cons(u, v), cons(w, z)) → LESS_LEAVES(concat(u, v), concat(w, z))
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(u, v)
LESS_LEAVES(cons(u, v), cons(w, z)) → CONCAT(w, z)

The TRS R consists of the following 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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

CONCAT(cons(u, v), y) → CONCAT(v, y)

The TRS R consists of the following 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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONCAT(cons(u, v), y) → CONCAT(v, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONCAT(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)
minus(x1, x2)  =  x1
0  =  0
s(x1)  =  s(x1)
quot(x1, x2)  =  x1
app(x1, x2)  =  app(x2)
nil  =  nil
add(x1, x2)  =  add
reverse(x1)  =  reverse(x1)
shuffle(x1)  =  shuffle(x1)
concat(x1, x2)  =  concat(x1, x2)
leaf  =  leaf
less_leaves(x1, x2)  =  less_leaves
false  =  false
true  =  true

Lexicographic Path Order [LPO].
Precedence:

[cons₂, concat₂] > less_leaves > [0, nil, add, false, true]
s₁ > [0, nil, add, false, true]
reverse₁ > app₁ > [0, nil, add, false, true]
shuffle₁ > [0, nil, add, false, true]
leaf > [0, nil, add, false, true]

The following usable rules [FROCOS05] were oriented:

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))

(9) Obligation:

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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:

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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(13) 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:

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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(14) 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:

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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(15) 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:

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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(16) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following 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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  x2
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0
quot(x1, x2)  =  x1
app(x1, x2)  =  app(x2)
nil  =  nil
add(x1, x2)  =  add
reverse(x1)  =  reverse(x1)
shuffle(x1)  =  shuffle(x1)
concat(x1, x2)  =  concat(x2)
leaf  =  leaf
cons(x1, x2)  =  x2
less_leaves(x1, x2)  =  less_leaves
false  =  false
true  =  true

Lexicographic Path Order [LPO].
Precedence:

[s₁, 0] > [nil, add, false, true]
reverse₁ > app₁ > [nil, add, false, true]
shuffle₁ > [nil, add, false, true]
concat₁ > [nil, add, false, true]
leaf > [nil, add, false, true]
less_leaves > [nil, add, false, true]

The following usable rules [FROCOS05] were oriented:

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))

(18) Obligation:

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following 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 set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0
quot(x1, x2)  =  x1
app(x1, x2)  =  app(x2)
nil  =  nil
add(x1, x2)  =  add
reverse(x1)  =  reverse(x1)
shuffle(x1)  =  shuffle(x1)
concat(x1, x2)  =  concat(x2)
leaf  =  leaf
cons(x1, x2)  =  x2
less_leaves(x1, x2)  =  less_leaves
false  =  false
true  =  true

Lexicographic Path Order [LPO].
Precedence:

[QUOT₂, s₁] > [nil, add, false, true]
0 > [nil, add, false, true]
reverse₁ > app₁ > [nil, add, false, true]
shuffle₁ > [nil, add, false, true]
concat₁ > [nil, add, false, true]
leaf > [nil, add, false, true]
less_leaves > [nil, add, false, true]

The following usable rules [FROCOS05] were oriented:

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))

(23) Obligation:

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
reverse(nil)
reverse(add(x0, x1))
shuffle(nil)
shuffle(add(x0, x1))
concat(leaf, x0)
concat(cons(x0, x1), x2)
less_leaves(x0, leaf)
less_leaves(leaf, cons(x0, x1))
less_leaves(cons(x0, x1), cons(x2, x3))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE