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:

:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

:12(x, :2(y, i1(x))) -> I1(y)
:12(e, x) -> I1(x)
:12(:2(x, y), z) -> I1(y)
:12(:2(x, y), z) -> :12(z, i1(y))
:12(x, :2(y, :2(i1(x), z))) -> I1(z)
:12(i1(x), :2(y, x)) -> I1(y)
:12(i1(x), :2(y, :2(x, z))) -> I1(z)
:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))
I1(:2(x, y)) -> :12(y, x)
:12(x, :2(y, :2(i1(x), z))) -> :12(i1(z), y)
:12(i1(x), :2(y, :2(x, z))) -> :12(i1(z), y)

The TRS R consists of the following rules:

:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(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:

:12(x, :2(y, i1(x))) -> I1(y)
:12(e, x) -> I1(x)
:12(:2(x, y), z) -> I1(y)
:12(:2(x, y), z) -> :12(z, i1(y))
:12(x, :2(y, :2(i1(x), z))) -> I1(z)
:12(i1(x), :2(y, x)) -> I1(y)
:12(i1(x), :2(y, :2(x, z))) -> I1(z)
:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))
I1(:2(x, y)) -> :12(y, x)
:12(x, :2(y, :2(i1(x), z))) -> :12(i1(z), y)
:12(i1(x), :2(y, :2(x, z))) -> :12(i1(z), y)

The TRS R consists of the following rules:

:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


:12(x, :2(y, i1(x))) -> I1(y)
:12(e, x) -> I1(x)
:12(:2(x, y), z) -> I1(y)
:12(:2(x, y), z) -> :12(z, i1(y))
:12(x, :2(y, :2(i1(x), z))) -> I1(z)
:12(i1(x), :2(y, x)) -> I1(y)
:12(i1(x), :2(y, :2(x, z))) -> I1(z)
I1(:2(x, y)) -> :12(y, x)
:12(x, :2(y, :2(i1(x), z))) -> :12(i1(z), y)
:12(i1(x), :2(y, :2(x, z))) -> :12(i1(z), y)
The remaining pairs can at least be oriented weakly.

:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( e ) = 2


POL( i1(x1) ) = x1


POL( :12(x1, x2) ) = max{0, 2x1 + 2x2 - 3}


POL( I1(x1) ) = max{0, 2x1 - 2}


POL( :2(x1, x2) ) = x1 + x2 + 3



The following usable rules [14] were oriented:

i1(e) -> e
:2(x, e) -> x
:2(e, x) -> i1(x)
:2(i1(x), :2(y, x)) -> i1(y)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)
i1(:2(x, y)) -> :2(y, x)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(x, :2(y, i1(x))) -> i1(y)
i1(i1(x)) -> x
:2(x, x) -> e



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

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

:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))

The TRS R consists of the following rules:

:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(z), y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


:12(:2(x, y), z) -> :12(x, :2(z, i1(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( e ) = 2


POL( i1(x1) ) = max{0, 3x1 - 2}


POL( :12(x1, x2) ) = 3x1 + 3


POL( :2(x1, x2) ) = 2x1 + 1



The following usable rules [14] 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:

:2(x, x) -> e
:2(x, e) -> x
i1(:2(x, y)) -> :2(y, x)
:2(:2(x, y), z) -> :2(x, :2(z, i1(y)))
:2(e, x) -> i1(x)
i1(i1(x)) -> x
i1(e) -> e
:2(x, :2(y, i1(x))) -> i1(y)
:2(x, :2(y, :2(i1(x), z))) -> :2(i1(z), y)
:2(i1(x), :2(y, x)) -> i1(y)
:2(i1(x), :2(y, :2(x, z))) -> :2(i1(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.