Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 1 + x₂   
POL(REMOVE(x₁, x₂)) = x₂   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 1 + x₁ + x₂   
POL(=(x₁, x₂)) = 0   
POL(PURGE(x₁)) = x₁   
POL(if(x₁, x₂, x₃)) = 0   
POL(nil) = 0   
POL(remove(x₁, x₂)) = 0   

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE