Termination w.r.t. Q proof of /tmp/tmpMM6GvD/revlist.trs

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:

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

↳ 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	/

/	0	0	\
\	0	0	/

x₁

/	1	0	\
\	0	0	/

x₂

M( rev2(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	0	\
\	0	0	/

x₂

M( s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( 0 ) =

/	0	\
\	0	/

M( rev1(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( nil ) =

/	0	\
\	0	/

M( rev(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

Tuple symbols:

M( REV2(x₁, x₂) ) =

[

]

x₁

[

]

x₂

M( REV(x₁) ) =

[

]

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.