Termination w.r.t. Q proof of TRS/secret05aprove3.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:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
FUNCTION(plus, dummy, x, y) → FUNCTION(iszero, x, x, x)
FUNCTION(if, false, x, y) → FUNCTION(p, x, x, y)
FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)
FUNCTION(if, false, x, y) → FUNCTION(third, x, y, y)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
FUNCTION(plus, dummy, x, y) → FUNCTION(iszero, x, x, x)
FUNCTION(if, false, x, y) → FUNCTION(p, x, x, y)
FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)
FUNCTION(if, false, x, y) → FUNCTION(third, x, y, y)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(plus, dummy, x, y) → FUNCTION(iszero, x, x, x)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
FUNCTION(if, false, x, y) → FUNCTION(p, x, x, y)
FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)
FUNCTION(if, false, x, y) → FUNCTION(third, x, y, y)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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.

FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FUNCTION(x1, x2, x3, x4) = x2
p = p
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

p > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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