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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

*1(2, 2) → .1(1, 0)
+1(3, x) → .1(1, +(min, x))
*1(+(y, z), x) → +1(*(x, y), *(x, z))
*1(.(x, y), z) → *1(y, z)
+1(3, x) → +1(min, x)
+1(.(x, y), z) → .1(x, +(y, z))
.1(x, .(y, z)) → .1(+(x, y), z)
.1(x, min) → .1(+(min, x), 3)
*1(2, min) → .1(min, 2)
.1(x, .(y, z)) → +1(x, y)
+1(*(2, x), x) → *1(3, x)
*1(.(x, y), z) → .1(*(x, z), *(y, z))
*1(3, x) → *1(min, x)
*1(+(y, z), x) → *1(x, y)
+1(x, x) → *1(2, x)
*1(3, x) → .1(x, *(min, x))
.1(x, min) → +1(min, x)
+1(.(x, y), z) → +1(y, z)
*1(.(x, y), z) → *1(x, z)
*1(+(y, z), x) → *1(x, 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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

*1(2, 2) → .1(1, 0)
+1(3, x) → .1(1, +(min, x))
*1(+(y, z), x) → +1(*(x, y), *(x, z))
+1(3, x) → +1(min, x)
*1(.(x, y), z) → *1(y, z)
+1(.(x, y), z) → .1(x, +(y, z))
*1(2, min) → .1(min, 2)
.1(x, min) → .1(+(min, x), 3)
.1(x, .(y, z)) → .1(+(x, y), z)
.1(x, .(y, z)) → +1(x, y)
+1(*(2, x), x) → *1(3, x)
*1(.(x, y), z) → .1(*(x, z), *(y, z))
*1(3, x) → *1(min, x)
*1(+(y, z), x) → *1(x, y)
+1(x, x) → *1(2, x)
*1(3, x) → .1(x, *(min, x))
+1(.(x, y), z) → +1(y, z)
.1(x, min) → +1(min, x)
*1(.(x, y), z) → *1(x, z)
*1(+(y, z), x) → *1(x, 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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 9 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

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

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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

*1(+(y, z), x) → *1(x, y)
*1(.(x, y), z) → *1(y, z)
*1(.(x, y), z) → *1(x, z)
*1(+(y, z), x) → *1(x, 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*1(+(y, z), x) → *1(x, y)
*1(.(x, y), z) → *1(y, z)
*1(.(x, y), z) → *1(x, z)
*1(+(y, z), x) → *1(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
+(x1, x2)  =  +(x1, x2)
.(x1, x2)  =  .(x1, x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.