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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x2)

From the DPs we obtained the following set of size-change graphs:

• REMOVE(x, .(y, z)) → REMOVE(x, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
remove(x1, x2)  =  x2
nil  =  nil
.(x1, x2)  =  .(x2)
if(x1, x2, x3)  =  x3
=(x1, x2)  =  =(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

.₁ > nil
=₂ > nil

Status:

nil: multiset
.₁: multiset
=₂: multiset

AFS:
remove(x1, x2)  =  x2
nil  =  nil
.(x1, x2)  =  .(x2)
if(x1, x2, x3)  =  x3
=(x1, x2)  =  =(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• PURGE(.(x, y)) → PURGE(remove(x, y)) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].

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

(10) TRUE