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 Order [17,21] with Interpretation:

POL( REMOVE2(x1, x2) ) = x1 + 2x2 + 3


POL( .2(x1, x2) ) = 3x1 + x2 + 1



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 Order [17,21] with Interpretation:

POL( PURGE1(x1) ) = x1


POL( =2(x1, x2) ) = 3x2 + 1


POL( if3(x1, ..., x3) ) = max{0, x1 + x2 - 3}


POL( .2(x1, x2) ) = 3x1 + 3x2 + 1


POL( nil ) = 1


POL( remove2(x1, x2) ) = 3x1 + 2x2



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.