Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

COUNT2(n, x) -> ISEMPTY1(left1(n))
NROFNODES1(n) -> COUNT2(n, 0)
COUNT2(n, x) -> LEFT1(n)
IF6(false, false, n, m, x, y) -> COUNT2(m, x)
COUNT2(n, x) -> INC1(x)
COUNT2(n, x) -> LEFT1(left1(n))
COUNT2(n, x) -> ISEMPTY1(n)
IF6(false, true, n, m, x, y) -> COUNT2(n, y)
COUNT2(n, x) -> IF6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
COUNT2(n, x) -> RIGHT1(n)
COUNT2(n, x) -> RIGHT1(left1(n))
INC1(s1(x)) -> INC1(x)

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

COUNT2(n, x) -> ISEMPTY1(left1(n))
NROFNODES1(n) -> COUNT2(n, 0)
COUNT2(n, x) -> LEFT1(n)
IF6(false, false, n, m, x, y) -> COUNT2(m, x)
COUNT2(n, x) -> INC1(x)
COUNT2(n, x) -> LEFT1(left1(n))
COUNT2(n, x) -> ISEMPTY1(n)
IF6(false, true, n, m, x, y) -> COUNT2(n, y)
COUNT2(n, x) -> IF6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
COUNT2(n, x) -> RIGHT1(n)
COUNT2(n, x) -> RIGHT1(left1(n))
INC1(s1(x)) -> INC1(x)

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 8 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP

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

INC1(s1(x)) -> INC1(x)

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


INC1(s1(x)) -> INC1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( INC1(x1) ) = x1


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

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

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP

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

IF6(false, false, n, m, x, y) -> COUNT2(m, x)
IF6(false, true, n, m, x, y) -> COUNT2(n, y)
COUNT2(n, x) -> IF6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))

The TRS R consists of the following rules:

isEmpty1(empty) -> true
isEmpty1(node2(l, r)) -> false
left1(empty) -> empty
left1(node2(l, r)) -> l
right1(empty) -> empty
right1(node2(l, r)) -> r
inc1(0) -> s1(0)
inc1(s1(x)) -> s1(inc1(x))
count2(n, x) -> if6(isEmpty1(n), isEmpty1(left1(n)), right1(n), node2(left1(left1(n)), node2(right1(left1(n)), right1(n))), x, inc1(x))
if6(true, b, n, m, x, y) -> x
if6(false, false, n, m, x, y) -> count2(m, x)
if6(false, true, n, m, x, y) -> count2(n, y)
nrOfNodes1(n) -> count2(n, 0)

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