Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/AProVESLASHLiveness6.4.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:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.

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

TOP2₂(x, free₁(y)) -> TOP1₂(x, check₁(new₁(y)))
TOP2₂(x, free₁(y)) -> CHECK₁(new₁(y))
TOP1₂(free₁(x), y) -> CHECK₁(new₁(x))
TOP2₂(x, free₁(y)) -> TOP1₂(new₁(x), check₁(y))
TOP2₂(x, free₁(y)) -> TOP1₂(check₁(x), new₁(y))
TOP2₂(x, free₁(y)) -> CHECK₁(x)
TOP2₂(x, free₁(y)) -> TOP1₂(check₁(new₁(x)), y)
CHECK₁(free₁(x)) -> CHECK₁(x)
CHECK₁(old₁(x)) -> OLD₁(check₁(x))
TOP1₂(free₁(x), y) -> CHECK₁(new₁(y))
TOP1₂(free₁(x), y) -> CHECK₁(x)
TOP1₂(free₁(x), y) -> NEW₁(x)
CHECK₁(new₁(x)) -> CHECK₁(x)
TOP2₂(x, free₁(y)) -> NEW₁(x)
TOP1₂(free₁(x), y) -> TOP2₂(check₁(x), new₁(y))
TOP1₂(free₁(x), y) -> TOP2₂(new₁(x), check₁(y))
TOP2₂(x, free₁(y)) -> CHECK₁(new₁(x))
TOP1₂(free₁(x), y) -> NEW₁(y)
TOP1₂(free₁(x), y) -> CHECK₁(y)
TOP1₂(free₁(x), y) -> TOP2₂(check₁(new₁(x)), y)
OLD₁(free₁(x)) -> OLD₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)
CHECK₁(new₁(x)) -> NEW₁(check₁(x))
TOP2₂(x, free₁(y)) -> NEW₁(y)
TOP2₂(x, free₁(y)) -> CHECK₁(y)
TOP1₂(free₁(x), y) -> TOP2₂(x, check₁(new₁(y)))
NEW₁(free₁(x)) -> NEW₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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:

TOP2₂(x, free₁(y)) -> TOP1₂(x, check₁(new₁(y)))
TOP2₂(x, free₁(y)) -> CHECK₁(new₁(y))
TOP1₂(free₁(x), y) -> CHECK₁(new₁(x))
TOP2₂(x, free₁(y)) -> TOP1₂(new₁(x), check₁(y))
TOP2₂(x, free₁(y)) -> TOP1₂(check₁(x), new₁(y))
TOP2₂(x, free₁(y)) -> CHECK₁(x)
TOP2₂(x, free₁(y)) -> TOP1₂(check₁(new₁(x)), y)
CHECK₁(free₁(x)) -> CHECK₁(x)
CHECK₁(old₁(x)) -> OLD₁(check₁(x))
TOP1₂(free₁(x), y) -> CHECK₁(new₁(y))
TOP1₂(free₁(x), y) -> CHECK₁(x)
TOP1₂(free₁(x), y) -> NEW₁(x)
CHECK₁(new₁(x)) -> CHECK₁(x)
TOP2₂(x, free₁(y)) -> NEW₁(x)
TOP1₂(free₁(x), y) -> TOP2₂(check₁(x), new₁(y))
TOP1₂(free₁(x), y) -> TOP2₂(new₁(x), check₁(y))
TOP2₂(x, free₁(y)) -> CHECK₁(new₁(x))
TOP1₂(free₁(x), y) -> NEW₁(y)
TOP1₂(free₁(x), y) -> CHECK₁(y)
TOP1₂(free₁(x), y) -> TOP2₂(check₁(new₁(x)), y)
OLD₁(free₁(x)) -> OLD₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)
CHECK₁(new₁(x)) -> NEW₁(check₁(x))
TOP2₂(x, free₁(y)) -> NEW₁(y)
TOP2₂(x, free₁(y)) -> CHECK₁(y)
TOP1₂(free₁(x), y) -> TOP2₂(x, check₁(new₁(y)))
NEW₁(free₁(x)) -> NEW₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 14 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

OLD₁(free₁(x)) -> OLD₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
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.

OLD₁(free₁(x)) -> OLD₁(x)

Used argument filtering: OLD₁(x1) = x1
free₁(x1) = free₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

NEW₁(free₁(x)) -> NEW₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
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.

NEW₁(free₁(x)) -> NEW₁(x)

Used argument filtering: NEW₁(x1) = x1
free₁(x1) = free₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

CHECK₁(new₁(x)) -> CHECK₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)
CHECK₁(free₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
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.

CHECK₁(free₁(x)) -> CHECK₁(x)

Used argument filtering: CHECK₁(x1) = x1
new₁(x1) = x1
old₁(x1) = x1
free₁(x1) = free₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

CHECK₁(new₁(x)) -> CHECK₁(x)
CHECK₁(old₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
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.

CHECK₁(old₁(x)) -> CHECK₁(x)

Used argument filtering: CHECK₁(x1) = x1
new₁(x1) = x1
old₁(x1) = old₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

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

CHECK₁(new₁(x)) -> CHECK₁(x)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

Q is empty.
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.

CHECK₁(new₁(x)) -> CHECK₁(x)

Used argument filtering: CHECK₁(x1) = x1
new₁(x1) = new₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

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

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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

          ↳ QDP

          ↳ QDP

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

TOP2₂(x, free₁(y)) -> TOP1₂(x, check₁(new₁(y)))
TOP1₂(free₁(x), y) -> TOP2₂(check₁(new₁(x)), y)
TOP1₂(free₁(x), y) -> TOP2₂(x, check₁(new₁(y)))
TOP1₂(free₁(x), y) -> TOP2₂(check₁(x), new₁(y))
TOP1₂(free₁(x), y) -> TOP2₂(new₁(x), check₁(y))
TOP2₂(x, free₁(y)) -> TOP1₂(check₁(x), new₁(y))
TOP2₂(x, free₁(y)) -> TOP1₂(new₁(x), check₁(y))
TOP2₂(x, free₁(y)) -> TOP1₂(check₁(new₁(x)), y)

The TRS R consists of the following rules:

top1₂(free₁(x), y) -> top2₂(check₁(new₁(x)), y)
top1₂(free₁(x), y) -> top2₂(new₁(x), check₁(y))
top1₂(free₁(x), y) -> top2₂(check₁(x), new₁(y))
top1₂(free₁(x), y) -> top2₂(x, check₁(new₁(y)))
top2₂(x, free₁(y)) -> top1₂(check₁(new₁(x)), y)
top2₂(x, free₁(y)) -> top1₂(new₁(x), check₁(y))
top2₂(x, free₁(y)) -> top1₂(check₁(x), new₁(y))
top2₂(x, free₁(y)) -> top1₂(x, check₁(new₁(y)))
new₁(free₁(x)) -> free₁(new₁(x))
old₁(free₁(x)) -> free₁(old₁(x))
new₁(serve) -> free₁(serve)
old₁(serve) -> free₁(serve)
check₁(free₁(x)) -> free₁(check₁(x))
check₁(new₁(x)) -> new₁(check₁(x))
check₁(old₁(x)) -> old₁(check₁(x))
check₁(old₁(x)) -> old₁(x)

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