Termination w.r.t. Q proof of /tmp/tmpXSfZ9Z/2.43.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:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

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:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) → IF(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) → IF(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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:

++¹(.(x, y), z) → ++¹(y, z)

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

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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.

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

↳ 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:

++¹(.(x, y), z) → ++¹(y, z)

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:

++¹(.(x, y), z) → ++¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

The TRS R consists of the following rules:

merge(nil, y) → y
merge(x, nil) → x
merge(.(x, y), .(u, v)) → if(<(x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
if(true, x, y) → x
if(false, x, y) → x

The set Q consists of the following terms:

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

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

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

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.

merge(nil, x0)
merge(x0, nil)
merge(.(x0, x1), .(x2, x3))
++(nil, x0)
++(.(x0, x1), x2)
if(true, x0, x1)
if(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))

From the DPs we obtained the following set of size-change graphs:

MERGE(.(x, y), .(u, v)) → MERGE(.(x, y), v)
The graph contains the following edges 1 >= 1, 2 > 2
MERGE(.(x, y), .(u, v)) → MERGE(y, .(u, v))
The graph contains the following edges 1 > 1, 2 >= 2