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) 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, z)
*¹(+(x, y), z) → *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)  =  *¹(x1)
*(x1, x2)  =  *(x2)
otimes(x1, x2)  =  otimes
+(x1, x2)  =  +(x1, x2)
oplus(x1, x2)  =  oplus
1  =  1

Recursive Path Order [RPO].
Precedence:

[*₁, +₂] > [*^1₁, otimes, oplus]
1 > [*^1₁, otimes, oplus]

The following usable rules [FROCOS05] were oriented:

*(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))

(4) Obligation:

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

*¹(x, *(y, z)) → *¹(otimes(x, 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.

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

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

Recursive Path Order [RPO].
Precedence:

*₂ > [*^1₁, otimes, oplus₂, +₂]
1 > [*^1₁, otimes, oplus₂, +₂]

The following usable rules [FROCOS05] were oriented:

*(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))

(6) Obligation:

Q DP problem:
P is empty.
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.

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE