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:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PURGE(.(x, y)) → REMOVE(x, y)
REMOVE(x, .(y, z)) → REMOVE(x, z)
PURGE(.(x, y)) → PURGE(remove(x, y))

The TRS R consists of the following rules:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(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:

PURGE(.(x, y)) → REMOVE(x, y)
REMOVE(x, .(y, z)) → REMOVE(x, z)
PURGE(.(x, y)) → PURGE(remove(x, y))

The TRS R consists of the following rules:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

REMOVE(x, .(y, z)) → REMOVE(x, z)

The TRS R consists of the following rules:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


REMOVE(x, .(y, z)) → REMOVE(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(REMOVE(x1, x2)) = (4)x_2   
POL(.(x1, x2)) = 1 + (4)x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(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:

PURGE(.(x, y)) → PURGE(remove(x, y))

The TRS R consists of the following rules:

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


PURGE(.(x, y)) → PURGE(remove(x, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(remove(x1, x2)) = 1 + x_1   
POL(if(x1, x2, x3)) = (3)x_1   
POL(=(x1, x2)) = 0   
POL(.(x1, x2)) = 4 + x_1 + x_2   
POL(nil) = 1   
POL(PURGE(x1)) = (3)x_1   
The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented:

remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(x, z)))



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

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

purge(nil) → nil
purge(.(x, y)) → .(x, purge(remove(x, y)))
remove(x, nil) → nil
remove(x, .(y, z)) → if(=(x, y), remove(x, z), .(y, remove(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.