Termination w.r.t. Q proof of /tmp/tmpEsHv70/cime1.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

Q is empty.

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

*¹(.(x, y), z) → .¹(*(x, z), *(y, z))
*¹(+(y, z), x) → *¹(x, y)
+¹(*(2, x), x) → *¹(3, x)
+¹(.(x, y), z) → .¹(x, +(y, z))
*¹(.(x, y), z) → *¹(x, z)
+¹(.(x, y), z) → +¹(y, z)
*¹(2, 2) → .¹(1, 0)
.¹(x, min) → +¹(min, x)
+¹(x, x) → *¹(2, x)
.¹(x, .(y, z)) → .¹(+(x, y), z)
*¹(.(x, y), z) → *¹(y, z)
*¹(2, min) → .¹(min, 2)
+¹(3, x) → .¹(1, +(min, x))
+¹(3, x) → +¹(min, x)
*¹(+(y, z), x) → +¹(*(x, y), *(x, z))
*¹(3, x) → *¹(min, x)
*¹(+(y, z), x) → *¹(x, z)
*¹(3, x) → .¹(x, *(min, x))
.¹(x, min) → .¹(+(min, x), 3)
.¹(x, .(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

*¹(.(x, y), z) → .¹(*(x, z), *(y, z))
*¹(+(y, z), x) → *¹(x, y)
+¹(*(2, x), x) → *¹(3, x)
+¹(.(x, y), z) → .¹(x, +(y, z))
*¹(.(x, y), z) → *¹(x, z)
+¹(.(x, y), z) → +¹(y, z)
*¹(2, 2) → .¹(1, 0)
.¹(x, min) → +¹(min, x)
+¹(x, x) → *¹(2, x)
.¹(x, .(y, z)) → .¹(+(x, y), z)
*¹(.(x, y), z) → *¹(y, z)
*¹(2, min) → .¹(min, 2)
+¹(3, x) → .¹(1, +(min, x))
+¹(3, x) → +¹(min, x)
*¹(+(y, z), x) → +¹(*(x, y), *(x, z))
*¹(3, x) → *¹(min, x)
*¹(+(y, z), x) → *¹(x, z)
*¹(3, x) → .¹(x, *(min, x))
.¹(x, min) → .¹(+(min, x), 3)
.¹(x, .(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ Narrowing

          ↳ QDP

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

+¹(*(2, x), x) → *¹(3, x)
+¹(.(x, y), z) → .¹(x, +(y, z))
+¹(3, x) → .¹(1, +(min, x))
+¹(.(x, y), z) → +¹(y, z)
.¹(x, .(y, z)) → .¹(+(x, y), z)
*¹(3, x) → .¹(x, *(min, x))
.¹(x, .(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule +¹(3, x) → .¹(1, +(min, x)) at position [1] we obtained the following new rules:

+¹(3, min) → .¹(1, *(2, min))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

          ↳ QDP

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

+¹(.(x, y), z) → .¹(x, +(y, z))
+¹(*(2, x), x) → *¹(3, x)
+¹(.(x, y), z) → +¹(y, z)
.¹(x, .(y, z)) → .¹(+(x, y), z)
*¹(3, x) → .¹(x, *(min, x))
+¹(3, min) → .¹(1, *(2, min))
.¹(x, .(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule +¹(.(x, y), z) → .¹(x, +(y, z)) at position [1] we obtained the following new rules:

+¹(.(y0, 1), 2) → .¹(y0, 3)
+¹(.(y0, 2), min) → .¹(y0, 1)
+¹(.(y0, *(2, x0)), *(min, x0)) → .¹(y0, x0)
+¹(.(y0, *(min, x0)), x0) → .¹(y0, 0)
+¹(.(y0, 1), min) → .¹(y0, 0)
+¹(.(y0, 3), x0) → .¹(y0, .(1, +(min, x0)))
+¹(.(y0, *(2, x0)), x0) → .¹(y0, *(3, x0))
+¹(.(y0, .(x0, x1)), x2) → .¹(y0, .(x0, +(x1, x2)))
+¹(.(y0, 0), x0) → .¹(y0, x0)
+¹(.(y0, x0), x0) → .¹(y0, *(2, x0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

          ↳ QDP

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

+¹(*(2, x), x) → *¹(3, x)
+¹(.(y0, 1), 2) → .¹(y0, 3)
+¹(.(y0, 2), min) → .¹(y0, 1)
+¹(.(x, y), z) → +¹(y, z)
+¹(.(y0, *(2, x0)), *(min, x0)) → .¹(y0, x0)
+¹(.(y0, 1), min) → .¹(y0, 0)
.¹(x, .(y, z)) → .¹(+(x, y), z)
+¹(.(y0, *(2, x0)), x0) → .¹(y0, *(3, x0))
+¹(.(y0, 0), x0) → .¹(y0, x0)
+¹(.(y0, *(min, x0)), x0) → .¹(y0, 0)
+¹(.(y0, 3), x0) → .¹(y0, .(1, +(min, x0)))
+¹(.(y0, .(x0, x1)), x2) → .¹(y0, .(x0, +(x1, x2)))
*¹(3, x) → .¹(x, *(min, x))
+¹(.(y0, x0), x0) → .¹(y0, *(2, x0))
+¹(3, min) → .¹(1, *(2, min))
.¹(x, .(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

          ↳ QDP

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

+¹(*(2, x), x) → *¹(3, x)
+¹(.(x, y), z) → +¹(y, z)
+¹(.(y0, *(2, x0)), *(min, x0)) → .¹(y0, x0)
+¹(.(y0, 3), x0) → .¹(y0, .(1, +(min, x0)))
.¹(x, .(y, z)) → .¹(+(x, y), z)
+¹(.(y0, .(x0, x1)), x2) → .¹(y0, .(x0, +(x1, x2)))
+¹(.(y0, *(2, x0)), x0) → .¹(y0, *(3, x0))
*¹(3, x) → .¹(x, *(min, x))
+¹(.(y0, x0), x0) → .¹(y0, *(2, x0))
+¹(.(y0, 0), x0) → .¹(y0, x0)
+¹(3, min) → .¹(1, *(2, min))
.¹(x, .(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

*¹(+(y, z), x) → *¹(x, y)
*¹(.(x, y), z) → *¹(x, z)
*¹(+(y, z), x) → *¹(x, z)
*¹(.(x, y), z) → *¹(y, z)

The TRS R consists of the following rules:

*(0, x) → 0
*(1, x) → x
*(2, 2) → .(1, 0)
*(3, x) → .(x, *(min, x))
*(min, min) → 1
*(2, min) → .(min, 2)
*(.(x, y), z) → .(*(x, z), *(y, z))
*(+(y, z), x) → +(*(x, y), *(x, z))
+(0, x) → x
+(x, x) → *(2, x)
+(1, 2) → 3
+(1, min) → 0
+(2, min) → 1
+(3, x) → .(1, +(min, x))
+(.(x, y), z) → .(x, +(y, z))
+(*(2, x), x) → *(3, x)
+(*(min, x), x) → 0
+(*(2, v), *(min, v)) → v
.(min, 3) → min
.(x, min) → .(+(min, x), 3)
.(0, x) → x
.(x, .(y, z)) → .(+(x, y), z)

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