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:

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

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:

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

The TRS R consists of the following rules:

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

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 1 less node.

(4) Complex Obligation (AND)

(5) 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, y), z)

The TRS R consists of the following rules:

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

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(x, y)
+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)
0  =  0
i(x1)  =  i

Recursive path order with status [RPO].
Precedence:

+^1₂ > +₂ > 0
i > 0

Status:

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

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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:

I(+(x, y)) → I(y)
I(+(x, y)) → I(x)

The TRS R consists of the following rules:

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

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.

I(+(x, y)) → I(y)
I(+(x, y)) → I(x)

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

+₂ > I₁

Status:

I₁: [1]
+₂: multiset

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE