(0) Obligation:

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

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

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:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(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:

REV1(X, cons(Y, L)) → REV1(Y, L)
REV(cons(X, L)) → REV1(X, L)
REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(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 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

REV1(X, cons(Y, L)) → REV1(Y, L)

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(8) 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.

(9) Obligation:

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

REV1(X, cons(Y, L)) → REV1(Y, L)

R is empty.
The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(10) 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].

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

(11) Obligation:

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

REV1(X, cons(Y, L)) → REV1(Y, L)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) 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:

  • REV1(X, cons(Y, L)) → REV1(Y, L)
    The graph contains the following edges 2 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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

REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(REV(x1)) = x1   
POL(REV2(x1, x2)) = x2   
POL(cons(x1, x2)) = 1 + x2   
POL(nil) = 1   
POL(rev(x1)) = x1   
POL(rev1(x1, x2)) = 1 + x2   
POL(rev2(x1, x2)) = x2   
POL(s(x1)) = 1   

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

(16) Obligation:

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

REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(18) TRUE