Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, z)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, 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:

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, z)))

The set Q consists of the following terms:

purge1(nil)
purge1(.2(x0, x1))
remove2(x0, nil)
remove2(x0, .2(x1, x2))


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

REMOVE2(x, .2(y, z)) -> REMOVE2(x, z)
PURGE1(.2(x, y)) -> REMOVE2(x, y)
PURGE1(.2(x, y)) -> PURGE1(remove2(x, y))

The TRS R consists of the following rules:

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, z)))

The set Q consists of the following terms:

purge1(nil)
purge1(.2(x0, x1))
remove2(x0, nil)
remove2(x0, .2(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:

REMOVE2(x, .2(y, z)) -> REMOVE2(x, z)
PURGE1(.2(x, y)) -> REMOVE2(x, y)
PURGE1(.2(x, y)) -> PURGE1(remove2(x, y))

The TRS R consists of the following rules:

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, z)))

The set Q consists of the following terms:

purge1(nil)
purge1(.2(x0, x1))
remove2(x0, nil)
remove2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

REMOVE2(x, .2(y, z)) -> REMOVE2(x, z)

The TRS R consists of the following rules:

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, z)))

The set Q consists of the following terms:

purge1(nil)
purge1(.2(x0, x1))
remove2(x0, nil)
remove2(x0, .2(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.

REMOVE2(x, .2(y, z)) -> REMOVE2(x, z)
Used argument filtering: REMOVE2(x1, x2)  =  x2
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ PisEmptyProof

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

purge1(nil) -> nil
purge1(.2(x, y)) -> .2(x, purge1(remove2(x, y)))
remove2(x, nil) -> nil
remove2(x, .2(y, z)) -> if3(=2(x, y), remove2(x, z), .2(y, remove2(x, z)))

The set Q consists of the following terms:

purge1(nil)
purge1(.2(x0, x1))
remove2(x0, nil)
remove2(x0, .2(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.