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

The TRS R consists of the following rules:

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

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)
oplus(x1, x2)  =  oplus(x1, x2)
otimes(x1, x2)  =  otimes

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

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

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

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

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

• *¹(+(x, y), z) → *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