Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

(6) Obligation:

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE