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, z)) → +(+(x, y), z)
+(*(x, y), +(x, 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, z)) → +¹(+(x, y), z)
+¹(x, +(y, z)) → +¹(x, y)
+¹(*(x, y), +(x, 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, z)) → +(+(x, y), z)
+(*(x, y), +(x, 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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(*(x₁, x₂)) = 1 + x₂   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(+¹(x₁, x₂)) = x₂   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(*(x₁, x₂)) = 0   
POL(+(x₁, x₂)) = 1 + x₁ + x₂   
POL(+¹(x₁, x₂)) = x₂   

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