Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/secret2005SLASHaprove3.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:

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

The set Q consists of the following terms:

function₄(iszero, 0, x0, x1)
function₄(iszero, s₁(x0), x1, x2)
function₄(p, 0, x0, x1)
function₄(p, s₁(0), x0, x1)
function₄(p, s₁(s₁(x0)), x1, x2)
function₄(plus, x0, x1, x2)
function₄(if, true, x0, x1)
function₄(if, false, x0, x1)
function₄(third, x0, x1, x2)

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

FUNCTION₄(plus, dummy, x, y) -> FUNCTION₄(if, function₄(iszero, x, x, x), x, y)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
FUNCTION₄(plus, dummy, x, y) -> FUNCTION₄(iszero, x, x, x)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(p, x, x, y)
FUNCTION₄(p, s₁(s₁(x)), dummy, dummy2) -> FUNCTION₄(p, s₁(x), x, x)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(third, x, y, y)

The TRS R consists of the following rules:

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

The set Q consists of the following terms:

function₄(iszero, 0, x0, x1)
function₄(iszero, s₁(x0), x1, x2)
function₄(p, 0, x0, x1)
function₄(p, s₁(0), x0, x1)
function₄(p, s₁(s₁(x0)), x1, x2)
function₄(plus, x0, x1, x2)
function₄(if, true, x0, x1)
function₄(if, false, x0, x1)
function₄(third, x0, x1, x2)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

FUNCTION₄(plus, dummy, x, y) -> FUNCTION₄(if, function₄(iszero, x, x, x), x, y)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
FUNCTION₄(plus, dummy, x, y) -> FUNCTION₄(iszero, x, x, x)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(p, x, x, y)
FUNCTION₄(p, s₁(s₁(x)), dummy, dummy2) -> FUNCTION₄(p, s₁(x), x, x)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(third, x, y, y)

The TRS R consists of the following rules:

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

The set Q consists of the following terms:

function₄(iszero, 0, x0, x1)
function₄(iszero, s₁(x0), x1, x2)
function₄(p, 0, x0, x1)
function₄(p, s₁(0), x0, x1)
function₄(p, s₁(s₁(x0)), x1, x2)
function₄(plus, x0, x1, x2)
function₄(if, true, x0, x1)
function₄(if, false, x0, x1)
function₄(third, x0, x1, x2)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

FUNCTION₄(p, s₁(s₁(x)), dummy, dummy2) -> FUNCTION₄(p, s₁(x), x, x)

The TRS R consists of the following rules:

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

The set Q consists of the following terms:

function₄(iszero, 0, x0, x1)
function₄(iszero, s₁(x0), x1, x2)
function₄(p, 0, x0, x1)
function₄(p, s₁(0), x0, x1)
function₄(p, s₁(s₁(x0)), x1, x2)
function₄(plus, x0, x1, x2)
function₄(if, true, x0, x1)
function₄(if, false, x0, x1)
function₄(third, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FUNCTION₄(p, s₁(s₁(x)), dummy, dummy2) -> FUNCTION₄(p, s₁(x), x, x)

Used argument filtering: FUNCTION₄(x1, x2, x3, x4) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

The set Q consists of the following terms:

function₄(iszero, 0, x0, x1)
function₄(iszero, s₁(x0), x1, x2)
function₄(p, 0, x0, x1)
function₄(p, s₁(0), x0, x1)
function₄(p, s₁(s₁(x0)), x1, x2)
function₄(plus, x0, x1, x2)
function₄(if, true, x0, x1)
function₄(if, false, x0, x1)
function₄(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

FUNCTION₄(plus, dummy, x, y) -> FUNCTION₄(if, function₄(iszero, x, x, x), x, y)
FUNCTION₄(if, false, x, y) -> FUNCTION₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))

The TRS R consists of the following rules:

function₄(iszero, 0, dummy, dummy2) -> true
function₄(iszero, s₁(x), dummy, dummy2) -> false
function₄(p, 0, dummy, dummy2) -> 0
function₄(p, s₁(0), dummy, dummy2) -> 0
function₄(p, s₁(s₁(x)), dummy, dummy2) -> s₁(function₄(p, s₁(x), x, x))
function₄(plus, dummy, x, y) -> function₄(if, function₄(iszero, x, x, x), x, y)
function₄(if, true, x, y) -> y
function₄(if, false, x, y) -> function₄(plus, function₄(third, x, y, y), function₄(p, x, x, y), s₁(y))
function₄(third, x, y, z) -> z

The set Q consists of the following terms:

function₄(iszero, 0, x0, x1)
function₄(iszero, s₁(x0), x1, x2)
function₄(p, 0, x0, x1)
function₄(p, s₁(0), x0, x1)
function₄(p, s₁(s₁(x0)), x1, x2)
function₄(plus, x0, x1, x2)
function₄(if, true, x0, x1)
function₄(if, false, x0, x1)
function₄(third, x0, x1, x2)

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