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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
*(x1, x2)  =  x2
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• +¹(x, +(y, z)) → +¹(+(x, y), z) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(x, +(y, z)) → +¹(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(*(x, y), +(x, z)) → +¹(y, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(*(x, y), +(*(x, z), u)) → +¹(*(x, +(y, z)), u) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(*(x, y), +(*(x, z), u)) → +¹(y, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(4) TRUE