Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ RRRPoloQTRSProof

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.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

:(x, x) → e
:(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)
Used ordering:
Polynomial interpretation [25]:

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




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ RuleRemovalProof

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

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

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.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: POLO with Polynomial interpretation [25]:

POL(:(x1, x2)) = 2 + x1 + x2   
POL(:1(x1, x2)) = 2·x1 + 2·x2   
POL(I(x1)) = 1 + 2·x1   
POL(e) = 1   
POL(i(x1)) = x1   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ QDPSizeChangeProof

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: