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:

isEmpty(empty) → true
isEmpty(node(l, r)) → false
left(empty) → empty
left(node(l, r)) → l
right(empty) → empty
right(node(l, r)) → r
inc(0) → s(0)
inc(s(x)) → s(inc(x))
count(n, x) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
if(true, b, n, m, x, y) → x
if(false, false, n, m, x, y) → count(m, x)
if(false, true, n, m, x, y) → count(n, y)
nrOfNodes(n) → count(n, 0)

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:

isEmpty(empty) → true
isEmpty(node(l, r)) → false
left(empty) → empty
left(node(l, r)) → l
right(empty) → empty
right(node(l, r)) → r
inc(0) → s(0)
inc(s(x)) → s(inc(x))
count(n, x) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
if(true, b, n, m, x, y) → x
if(false, false, n, m, x, y) → count(m, x)
if(false, true, n, m, x, y) → count(n, y)
nrOfNodes(n) → count(n, 0)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1))
left(empty)
left(node(x0, x1))
right(empty)
right(node(x0, x1))
inc(0)
inc(s(x0))
count(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
nrOfNodes(x0)

(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:

INC(s(x)) → INC(x)
COUNT(n, x) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
COUNT(n, x) → ISEMPTY(n)
COUNT(n, x) → ISEMPTY(left(n))
COUNT(n, x) → LEFT(n)
COUNT(n, x) → RIGHT(n)
COUNT(n, x) → LEFT(left(n))
COUNT(n, x) → RIGHT(left(n))
COUNT(n, x) → INC(x)
IF(false, false, n, m, x, y) → COUNT(m, x)
IF(false, true, n, m, x, y) → COUNT(n, y)
NROFNODES(n) → COUNT(n, 0)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, r)) → false
left(empty) → empty
left(node(l, r)) → l
right(empty) → empty
right(node(l, r)) → r
inc(0) → s(0)
inc(s(x)) → s(inc(x))
count(n, x) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
if(true, b, n, m, x, y) → x
if(false, false, n, m, x, y) → count(m, x)
if(false, true, n, m, x, y) → count(n, y)
nrOfNodes(n) → count(n, 0)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1))
left(empty)
left(node(x0, x1))
right(empty)
right(node(x0, x1))
inc(0)
inc(s(x0))
count(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
nrOfNodes(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, r)) → false
left(empty) → empty
left(node(l, r)) → l
right(empty) → empty
right(node(l, r)) → r
inc(0) → s(0)
inc(s(x)) → s(inc(x))
count(n, x) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
if(true, b, n, m, x, y) → x
if(false, false, n, m, x, y) → count(m, x)
if(false, true, n, m, x, y) → count(n, y)
nrOfNodes(n) → count(n, 0)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1))
left(empty)
left(node(x0, x1))
right(empty)
right(node(x0, x1))
inc(0)
inc(s(x0))
count(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
nrOfNodes(x0)

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.

INC(s(x)) → INC(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INC(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

isEmpty(empty) → true
isEmpty(node(l, r)) → false
left(empty) → empty
left(node(l, r)) → l
right(empty) → empty
right(node(l, r)) → r
inc(0) → s(0)
inc(s(x)) → s(inc(x))
count(n, x) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
if(true, b, n, m, x, y) → x
if(false, false, n, m, x, y) → count(m, x)
if(false, true, n, m, x, y) → count(n, y)
nrOfNodes(n) → count(n, 0)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1))
left(empty)
left(node(x0, x1))
right(empty)
right(node(x0, x1))
inc(0)
inc(s(x0))
count(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
nrOfNodes(x0)

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:

COUNT(n, x) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
IF(false, false, n, m, x, y) → COUNT(m, x)
IF(false, true, n, m, x, y) → COUNT(n, y)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, r)) → false
left(empty) → empty
left(node(l, r)) → l
right(empty) → empty
right(node(l, r)) → r
inc(0) → s(0)
inc(s(x)) → s(inc(x))
count(n, x) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), node(right(left(n)), right(n))), x, inc(x))
if(true, b, n, m, x, y) → x
if(false, false, n, m, x, y) → count(m, x)
if(false, true, n, m, x, y) → count(n, y)
nrOfNodes(n) → count(n, 0)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1))
left(empty)
left(node(x0, x1))
right(empty)
right(node(x0, x1))
inc(0)
inc(s(x0))
count(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
nrOfNodes(x0)

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