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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(:(x₁, x₂)) = 1 + x₁ + x₂   
POL(e) = 1   
POL(i(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

:(x, e) → x
:(e, x) → i(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)

(2) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(:(x₁, x₂)) = 2 + x₁ + x₂   
POL(e) = 0   
POL(i(x₁)) = x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

:(x, x) → e

(4) Obligation:

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

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

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) 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)

The TRS R consists of the following rules:

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

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

(7) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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

Used ordering: Polynomial interpretation [POLO]:

POL(:(x₁, x₂)) = 1 + x₁ + x₂   
POL(:¹(x₁, x₂)) = x₁ + x₂   
POL(I(x₁)) = x₁   
POL(e) = 0   
POL(i(x₁)) = x₁   

(8) 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:

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

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• :¹(:(x, y), z) → :¹(x, :(z, i(y)))
  The graph contains the following edges 1 > 1

(10) TRUE