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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE

(0) Obligation:

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

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(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(x, +(y, z)) → +1(*(x, y), *(x, z))
*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
*1(+(x, y), z) → +1(*(x, z), *(y, z))
*1(+(x, y), z) → *1(x, z)
*1(+(x, y), z) → *1(y, z)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) 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(y, z)
+1(+(x, y), z) → +1(x, +(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(0) = 0   
POL(i(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) 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

(12) Obligation:

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

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

The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) 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(x, y)
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

(14) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(x, 0) → x
+(x, i(x)) → 0
+(+(x, y), z) → +(x, +(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(+(x, y), z) → +(*(x, z), *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE