Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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

The TRS R consists of the following rules:

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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 1 SCC with 1 less node.

(4) Obligation:

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

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

The TRS R consists of the following rules:

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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

(5) 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(x, +(y, z))
+¹(*(x, y), *(x, z)) → +¹(y, z)
+¹(+(x, y), z) → +¹(y, z)
+¹(*(x, y), +(*(x, z), u)) → +¹(y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2)  =  x1
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x1, x2)
u  =  u

Lexicographic Path Order [LPO].
Precedence:

*₂ > +₂
u > +₂

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(*(x, y), *(x, z)) → *(x, +(y, z))
+(+(x, y), z) → +(x, +(y, z))
+(*(x, y), +(*(x, z), u)) → +(*(x, +(y, z)), u)

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE