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, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), 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:

I(:(x, y)) → :¹(y, x)
:¹(:(x, y), z) → :¹(x, :(z, i(y)))
:¹(:(x, y), z) → :¹(z, i(y))
:¹(:(x, y), z) → I(y)
:¹(e, x) → I(x)
:¹(x, :(y, i(x))) → I(y)
:¹(x, :(y, :(i(x), z))) → :¹(i(z), y)
:¹(x, :(y, :(i(x), z))) → I(z)
:¹(i(x), :(y, x)) → I(y)
:¹(i(x), :(y, :(x, z))) → :¹(i(z), y)
:¹(i(x), :(y, :(x, z))) → I(z)

The TRS R consists of the following rules:

:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), 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.

I(:(x, y)) → :¹(y, x)
:¹(:(x, y), z) → :¹(z, i(y))
:¹(:(x, y), z) → I(y)
:¹(e, x) → I(x)
:¹(x, :(y, i(x))) → I(y)
:¹(x, :(y, :(i(x), z))) → :¹(i(z), y)
:¹(x, :(y, :(i(x), z))) → I(z)
:¹(i(x), :(y, x)) → I(y)
:¹(i(x), :(y, :(x, z))) → :¹(i(z), y)
:¹(i(x), :(y, :(x, z))) → I(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 + x₂   
POL(I(x₁)) = x₁   
POL(e) = 1   
POL(i(x₁)) = x₁   

The following usable rules [FROCOS05] were oriented:

:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(:(x, y)) → :(y, x)
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), y)
i(i(x)) → x
i(e) → e
:(x, x) → e
:(x, e) → x

(4) Obligation:

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

:¹(:(x, y), z) → :¹(x, :(z, i(y)))

The TRS R consists of the following rules:

:(x, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), 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, :(z, i(y)))

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(e) = 0   
POL(i(x₁)) = 0   

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, x) → e
:(x, e) → x
i(:(x, y)) → :(y, x)
:(:(x, y), z) → :(x, :(z, i(y)))
:(e, x) → i(x)
i(i(x)) → x
i(e) → e
:(x, :(y, i(x))) → i(y)
:(x, :(y, :(i(x), z))) → :(i(z), y)
:(i(x), :(y, x)) → i(y)
:(i(x), :(y, :(x, z))) → :(i(z), 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