Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


QTRS
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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))


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

REV1(X, cons(Y, L)) → REV1(Y, L)
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV2(Y, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV(cons(X, L)) → REV1(X, 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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ UsableRulesProof
              ↳ QDP

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP

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.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
                ↳ QDPOrderProof

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

REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV2(Y, L)
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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
The remaining pairs can at least be oriented weakly.

REV(cons(X, L)) → REV2(X, L)
Used ordering: Polynomial interpretation [25]:

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

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ DependencyGraphProof
                ↳ QDPOrderProof

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

REV(cons(X, L)) → REV2(X, 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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
The remaining pairs can at least be oriented weakly.

REV(cons(X, L)) → REV2(X, L)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( cons(x1, x2) ) =
/1\
\0/
+
/00\
\00/
·x1+
/10\
\00/
·x2

M( rev2(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/10\
\00/
·x2

M( s(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( 0 ) =
/0\
\0/

M( rev1(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( nil ) =
/0\
\0/

M( rev(x1) ) =
/0\
\0/
+
/10\
\00/
·x1

Tuple symbols:
M( REV2(x1, x2) ) = 1+
[0,0]
·x1+
[1,0]
·x2

M( REV(x1) ) = 0+
[1,0]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                ↳ QDPOrderProof
QDP

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

REV(cons(X, L)) → REV2(X, 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.