Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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: Polynomial interpretation [POLO]:

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

The TRS R consists of the following rules:

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

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)
*1(x, +(y, z)) → *1(x, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

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

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