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:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

(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:

NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)
NTHTAIL(n, l) → GE(n, length(l))
NTHTAIL(n, l) → LENGTH(l)
COND(false, n, l) → TAIL(nthtail(s(n), l))
COND(false, n, l) → NTHTAIL(s(n), l)
LENGTH(cons(x, l)) → LENGTH(l)
GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.

GE(s(u), s(v)) → GE(u, v)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
GE(x1, x2)  =  GE(x1, x2)
s(x1)  =  s(x1)
nthtail(x1, x2)  =  x2
cond(x1, x2, x3)  =  x3
ge(x1, x2)  =  ge
length(x1)  =  x1
true  =  true
false  =  false
tail(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
0  =  0

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

GE₂ > [s₁, ge, true, false]
nil > 0 > [s₁, ge, true, false]
cons₂ > [s₁, ge, true, false]

Status:

GE₂: multiset
s₁: multiset
ge: multiset
true: multiset
false: multiset
nil: multiset
cons₂: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

(9) Obligation:

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

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

LENGTH(cons(x, l)) → LENGTH(l)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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.

LENGTH(cons(x, l)) → LENGTH(l)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LENGTH(x1)  =  x1
cons(x1, x2)  =  cons(x2)
nthtail(x1, x2)  =  nthtail(x1, x2)
cond(x1, x2, x3)  =  cond(x2, x3)
ge(x1, x2)  =  ge
length(x1)  =  length(x1)
true  =  true
false  =  false
tail(x1)  =  x1
s(x1)  =  s
nil  =  nil
0  =  0

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

cons₁ > [ge, length₁, true, false, s, 0]
[nthtail₂, cond₂] > [ge, length₁, true, false, s, 0]
nil > [ge, length₁, true, false, s, 0]

Status:

cons₁: multiset
nthtail₂: [1,2]
cond₂: [1,2]
ge: multiset
length₁: multiset
true: multiset
false: multiset
s: []
nil: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

(14) Obligation:

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

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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:

COND(false, n, l) → NTHTAIL(s(n), l)
NTHTAIL(n, l) → COND(ge(n, length(l)), n, l)

The TRS R consists of the following rules:

nthtail(n, l) → cond(ge(n, length(l)), n, l)
cond(true, n, l) → l
cond(false, n, l) → tail(nthtail(s(n), l))
tail(nil) → nil
tail(cons(x, l)) → l
length(nil) → 0
length(cons(x, l)) → s(length(l))
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)

The set Q consists of the following terms:

nthtail(x0, x1)
cond(true, x0, x1)
cond(false, x0, x1)
tail(nil)
tail(cons(x0, x1))
length(nil)
length(cons(x0, x1))
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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