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, 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: Combined order from the following AFS and order.
+¹(x1, x2)  =  +¹(x1, x2)
+(x1, x2)  =  +(x1, x2)
i(x1)  =  i
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

[i, 0] > [+^1₂, +₂]

Status:

i: multiset
0: multiset
+^1₂: [1,2]
+₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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

(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(x, y)

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)

Recursive path order with status [RPO].
Quasi-Precedence:

[*^1₁, +₂]

Status:

*^1₁: multiset
+₂: multiset

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(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.

(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(y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

[*^1₂, +₂]

Status:

*^1₂: multiset
+₂: multiset

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