(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) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) 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)))
The set Q consists of the following terms:
purge(nil)
purge(.(x0, x1))
remove(x0, nil)
remove(x0, .(x1, x2))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) 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)))
The set Q consists of the following terms:
purge(nil)
purge(.(x0, x1))
remove(x0, nil)
remove(x0, .(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(6) 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)))
The set Q consists of the following terms:
purge(nil)
purge(.(x0, x1))
remove(x0, nil)
remove(x0, .(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
(7) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REMOVE(x, .(y, z)) → REMOVE(x, z)
R is empty.
The set Q consists of the following terms:
purge(nil)
purge(.(x0, x1))
remove(x0, nil)
remove(x0, .(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
(9) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
purge(nil)
purge(.(x0, x1))
remove(x0, nil)
remove(x0, .(x1, x2))
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REMOVE(x, .(y, z)) → REMOVE(x, z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- REMOVE(x, .(y, z)) → REMOVE(x, z)
The graph contains the following edges 1 >= 1, 2 > 2
(12) YES