(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: SCNP Order with the following components:
Level mapping:
Top level AFS:
REMOVE(x0, x1, x2)  =  REMOVE(x0)

Tags:
REMOVE has argument tags [1,1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
REMOVE(x1, x2)  =  x2
.(x1, x2)  =  .(x1, x2)

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

Status:
.2: multiset


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: SCNP Order with the following components:
Level mapping:
Top level AFS:
PURGE(x0, x1)  =  PURGE(x1)

Tags:
PURGE has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
PURGE(x1)  =  PURGE
.(x1, x2)  =  .(x1, x2)
remove(x1, x2)  =  x1
nil  =  nil
if(x1, x2, x3)  =  if
=(x1, x2)  =  =(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
PURGE > [.2, nil, if, =2]

Status:
PURGE: multiset
.2: [1,2]
nil: multiset
if: multiset
=2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(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