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

The TRS R consists of the following rules:

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

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

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

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

The following usable rules [FROCOS05] were oriented:

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

(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(y, z)

The TRS R consists of the following rules:

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

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(y, z)

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

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

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

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