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
  ↳ 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)))

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:

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)))

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:

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)))

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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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)))

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.


REMOVE2(x, .2(y, z)) -> REMOVE2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 1 + 2·x2   
POL(REMOVE2(x1, x2)) = 2·x2   

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:

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.
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
            ↳ QDPOrderProof

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

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)))

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.


PURGE1(.2(x, y)) -> PURGE1(remove2(x, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 3 + 2·x1 + 3·x2   
POL(=2(x1, x2)) = 1   
POL(PURGE1(x1)) = 2·x1   
POL(if3(x1, x2, x3)) = 3 + x1   
POL(nil) = 1   
POL(remove2(x1, x2)) = 3·x2   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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)))

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.