Termination w.r.t. Q proof of TRS/secret06nrOfNodes.trs

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) ) = max{0, x₁ - 2}

POL( s₁(x₁) ) = x₁ + 3

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.