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
  ↳ DependencyPairsProof

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.

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

:1(i(x), :(y, :(x, z))) → :1(i(z), y)
:1(e, x) → I(x)
:1(i(x), :(y, :(x, z))) → I(z)
:1(:(x, y), z) → :1(z, i(y))
:1(:(x, y), z) → :1(x, :(z, i(y)))
:1(x, :(y, :(i(x), z))) → I(z)
:1(:(x, y), z) → I(y)
I(:(x, y)) → :1(y, x)
:1(i(x), :(y, x)) → I(y)
:1(x, :(y, i(x))) → I(y)
:1(x, :(y, :(i(x), z))) → :1(i(z), 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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


:1(i(x), :(y, :(x, z))) → :1(i(z), y)
:1(e, x) → I(x)
:1(i(x), :(y, :(x, z))) → I(z)
:1(:(x, y), z) → :1(z, i(y))
:1(x, :(y, :(i(x), z))) → I(z)
:1(:(x, y), z) → I(y)
I(:(x, y)) → :1(y, x)
:1(i(x), :(y, x)) → I(y)
:1(x, :(y, i(x))) → I(y)
:1(x, :(y, :(i(x), z))) → :1(i(z), y)
The remaining pairs can at least be oriented weakly.

:1(:(x, y), z) → :1(x, :(z, i(y)))
Used ordering: Polynomial interpretation [25,35]:

POL(i(x1)) = x_1   
POL(:(x1, x2)) = 2 + x_1 + x_2   
POL(:1(x1, x2)) = 1/4 + (2)x_1 + (2)x_2   
POL(e) = 0   
POL(I(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(i(x1)) = 0   
POL(:(x1, x2)) = 4 + (4)x_1 + (4)x_2   
POL(:1(x1, x2)) = (2)x_1   
POL(e) = 15/4   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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