Termination w.r.t. Q proof of /tmp/tmpte9G3d/aprove10.trs

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

0 QTRS

↳1 AAECC Innermost (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

(0) Obligation:

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

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)

The TRS R 2 is

a → c
a → d

The signature Sigma is {a, c, d}

(2) Obligation:

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

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

(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:

LESSELEMENTS(l, t) → LESSE(l, t, 0)
LESSE(l, t, n) → IF(le(length(l), n), le(length(toList(t)), n), l, t, n)
LESSE(l, t, n) → LE(length(l), n)
LESSE(l, t, n) → LENGTH(l)
LESSE(l, t, n) → LE(length(toList(t)), n)
LESSE(l, t, n) → LENGTH(toList(t))
LESSE(l, t, n) → TOLIST(t)
IF(false, false, l, t, n) → LESSE(l, t, s(n))
LENGTH(cons(n, l)) → LENGTH(l)
TOLIST(node(t1, n, t2)) → APPEND(toList(t1), cons(n, toList(t2)))
TOLIST(node(t1, n, t2)) → TOLIST(t1)
TOLIST(node(t1, n, t2)) → TOLIST(t2)
APPEND(cons(n, l1), l2) → APPEND(l1, l2)
LE(s(n), s(m)) → LE(n, m)

The TRS R consists of the following rules:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(n), s(m)) → LE(n, m)

The TRS R consists of the following rules:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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.

LE(s(n), s(m)) → LE(n, m)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = LE(x1)
s(x1) = s(x1)
lessElements(x1, x2) = lessElements(x1, x2)
lessE(x1, x2, x3) = lessE(x1, x2)
0 = 0
if(x1, x2, x3, x4, x5) = if(x3, x4)
le(x1, x2) = le(x1)
length(x1) = length(x1)
toList(x1) = toList(x1)
true = true
false = false
nil = nil
cons(x1, x2) = cons(x1, x2)
leaf = leaf
node(x1, x2, x3) = node(x1, x2, x3)
append(x1, x2) = append(x1, x2)
a = a
c = c
d = d

Lexicographic path order with status [LPO].
Quasi-Precedence:

[lessElements₂, lessE₂, if₂] > [le₁, true, false] > [LE₁, s₁]
[lessElements₂, lessE₂, if₂] > length₁ > [LE₁, s₁]
[lessElements₂, lessE₂, if₂] > length₁ > 0
toList₁ > nil
toList₁ > append₂ > cons₂ > [LE₁, s₁]
leaf > nil
node₃ > cons₂ > [LE₁, s₁]
[a, d] > c

Status:

LE₁: [1]
s₁: [1]
lessElements₂: [1,2]
lessE₂: [1,2]
0: []
if₂: [1,2]
le₁: [1]
length₁: [1]
toList₁: [1]
true: []
false: []
nil: []
cons₂: [1,2]
leaf: []
node₃: [1,3,2]
append₂: [1,2]
a: []
c: []
d: []

The following usable rules [FROCOS05] were oriented:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

(9) Obligation:

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

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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:

APPEND(cons(n, l1), l2) → APPEND(l1, l2)

The TRS R consists of the following rules:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

APPEND(cons(n, l1), l2) → APPEND(l1, l2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2) = APPEND(x1)
cons(x1, x2) = cons(x2)
lessElements(x1, x2) = lessElements(x1, x2)
lessE(x1, x2, x3) = lessE(x1, x2)
0 = 0
if(x1, x2, x3, x4, x5) = if(x3, x4)
le(x1, x2) = le
length(x1) = x1
toList(x1) = toList(x1)
true = true
false = false
s(x1) = s
nil = nil
leaf = leaf
node(x1, x2, x3) = node(x1, x3)
append(x1, x2) = append(x1, x2)
a = a
c = c
d = d

Lexicographic path order with status [LPO].
Quasi-Precedence:

APPEND₁ > s
[lessElements₂, lessE₂, if₂] > 0 > [le, true, false] > s
toList₁ > [nil, leaf] > 0 > [le, true, false] > s
toList₁ > append₂ > cons₁ > s
node₂ > append₂ > cons₁ > s
a > c > s
a > d > s

Status:

APPEND₁: [1]
cons₁: [1]
lessElements₂: [2,1]
lessE₂: [2,1]
0: []
if₂: [2,1]
le: []
toList₁: [1]
true: []
false: []
s: []
nil: []
leaf: []
node₂: [2,1]
append₂: [1,2]
a: []
c: []
d: []

The following usable rules [FROCOS05] were oriented:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

(14) Obligation:

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

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

TOLIST(node(t1, n, t2)) → TOLIST(t2)
TOLIST(node(t1, n, t2)) → TOLIST(t1)

The TRS R consists of the following rules:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOLIST(node(t1, n, t2)) → TOLIST(t2)
TOLIST(node(t1, n, t2)) → TOLIST(t1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOLIST(x1) = TOLIST(x1)
node(x1, x2, x3) = node(x1, x2, x3)
lessElements(x1, x2) = lessElements(x1, x2)
lessE(x1, x2, x3) = lessE(x1, x2)
0 = 0
if(x1, x2, x3, x4, x5) = if(x3, x4)
le(x1, x2) = le(x1, x2)
length(x1) = length(x1)
toList(x1) = toList(x1)
true = true
false = false
s(x1) = x1
nil = nil
cons(x1, x2) = x2
leaf = leaf
append(x1, x2) = append(x1, x2)
a = a
c = c
d = d

Lexicographic path order with status [LPO].
Quasi-Precedence:

TOLIST₁ > [toList₁, nil, leaf]
node₃ > append₂ > [toList₁, nil, leaf]
lessElements₂ > [lessE₂, if₂] > [toList₁, nil, leaf]
lessElements₂ > 0 > [toList₁, nil, leaf]
le₂ > true > [toList₁, nil, leaf]
le₂ > false > [lessE₂, if₂] > [toList₁, nil, leaf]
length₁ > 0 > [toList₁, nil, leaf]
a > c > [toList₁, nil, leaf]
a > d > [toList₁, nil, leaf]

Status:

TOLIST₁: [1]
node₃: [2,3,1]
lessElements₂: [2,1]
lessE₂: [2,1]
0: []
if₂: [2,1]
le₂: [1,2]
length₁: [1]
toList₁: [1]
true: []
false: []
nil: []
leaf: []
append₂: [1,2]
a: []
c: []
d: []

The following usable rules [FROCOS05] were oriented:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

(19) Obligation:

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

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

LENGTH(cons(n, l)) → LENGTH(l)

The TRS R consists of the following rules:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LENGTH(cons(n, l)) → LENGTH(l)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1) = LENGTH(x1)
cons(x1, x2) = cons(x2)
lessElements(x1, x2) = lessElements(x1, x2)
lessE(x1, x2, x3) = lessE(x1, x2, x3)
0 = 0
if(x1, x2, x3, x4, x5) = if(x3, x4, x5)
le(x1, x2) = le
length(x1) = length(x1)
toList(x1) = toList(x1)
true = true
false = false
s(x1) = s
nil = nil
leaf = leaf
node(x1, x2, x3) = node(x1, x3)
append(x1, x2) = append(x1, x2)
a = a
c = c
d = d

Lexicographic path order with status [LPO].
Quasi-Precedence:

lessElements₂ > [lessE₃, if₃] > length₁ > s
lessElements₂ > [0, nil] > s
[le, true, false] > [lessE₃, if₃] > length₁ > s
toList₁ > [0, nil] > s
toList₁ > append₂ > [LENGTH₁, cons₁] > length₁ > s
leaf > [0, nil] > s
node₂ > [LENGTH₁, cons₁] > length₁ > s
[a, c, d] > s

Status:

LENGTH₁: [1]
cons₁: [1]
lessElements₂: [2,1]
lessE₃: [1,2,3]
0: []
if₃: [1,2,3]
le: []
length₁: [1]
toList₁: [1]
true: []
false: []
s: []
nil: []
leaf: []
node₂: [2,1]
append₂: [2,1]
a: []
c: []
d: []

The following usable rules [FROCOS05] were oriented:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

(24) Obligation:

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

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

IF(false, false, l, t, n) → LESSE(l, t, s(n))
LESSE(l, t, n) → IF(le(length(l), n), le(length(toList(t)), n), l, t, n)

The TRS R consists of the following rules:

lessElements(l, t) → lessE(l, t, 0)
lessE(l, t, n) → if(le(length(l), n), le(length(toList(t)), n), l, t, n)
if(true, b, l, t, n) → l
if(false, true, l, t, n) → t
if(false, false, l, t, n) → lessE(l, t, s(n))
length(nil) → 0
length(cons(n, l)) → s(length(l))
toList(leaf) → nil
toList(node(t1, n, t2)) → append(toList(t1), cons(n, toList(t2)))
append(nil, l2) → l2
append(cons(n, l1), l2) → cons(n, append(l1, l2))
le(s(n), 0) → false
le(0, m) → true
le(s(n), s(m)) → le(n, m)
a → c
a → d

The set Q consists of the following terms:

lessElements(x0, x1)
lessE(x0, x1, x2)
if(true, x0, x1, x2, x3)
if(false, true, x0, x1, x2)
if(false, false, x0, x1, x2)
length(nil)
length(cons(x0, x1))
toList(leaf)
toList(node(x0, x1, x2))
append(nil, x0)
append(cons(x0, x1), x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
a

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