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, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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(x, y)
I(*(x, y)) → *¹(i(y), i(x))
I(*(x, y)) → I(y)
I(*(x, y)) → I(x)
*¹(*(i(x), k(y, z)), x) → K(*(*(i(x), y), x), *(*(i(x), z), x))
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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(x, y)
*¹(x, *(y, z)) → *¹(*(x, y), z)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), y), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), y)
*¹(*(i(x), k(y, z)), x) → *¹(*(i(x), z), x)
*¹(*(i(x), k(y, z)), x) → *¹(i(x), z)

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

i₁ > [*^1₂, *₂] > [k₂, 1]

Status:

*^1₂: [2,1]
*₂: [2,1]
i₁: [1]
k₂: [1,2]
1: []

The following usable rules [FROCOS05] were oriented:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

(7) Obligation:

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

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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(x)
I(*(x, y)) → I(y)

The TRS R consists of the following rules:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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(x)
I(*(x, y)) → I(y)

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

i₁ > [I₁, *₂, 1]

Status:

I₁: [1]
*₂: [2,1]
1: []
i₁: [1]

The following usable rules [FROCOS05] were oriented:

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

(12) Obligation:

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

*(x, 1) → x
*(1, y) → y
*(i(x), x) → 1
*(x, i(x)) → 1
*(x, *(y, z)) → *(*(x, y), z)
i(1) → 1
*(*(x, y), i(y)) → x
*(*(x, i(y)), y) → x
i(i(x)) → x
i(*(x, y)) → *(i(y), i(x))
k(x, 1) → 1
k(x, x) → 1
*(k(x, y), k(y, x)) → 1
*(*(i(x), k(y, z)), x) → k(*(*(i(x), y), x), *(*(i(x), z), x))
k(*(x, i(y)), *(y, i(x))) → 1

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