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(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

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

The TRS R consists of the following rules:

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

.12(.2(x, y), z) -> .12(y, z)
.12(.2(x, y), z) -> .12(x, .2(y, z))

The TRS R consists of the following rules:

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

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

POL(.2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(.12(x1, x2)) = 2·x1   
POL(1) = 0   
POL(i1(x1)) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

I1(.2(x, y)) -> I1(y)
I1(.2(x, y)) -> I1(x)

The TRS R consists of the following rules:

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

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.


I1(.2(x, y)) -> I1(y)
I1(.2(x, y)) -> I1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(I1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

.2(1, x) -> x
.2(x, 1) -> x
.2(i1(x), x) -> 1
.2(x, i1(x)) -> 1
i1(1) -> 1
i1(i1(x)) -> x
.2(i1(y), .2(y, z)) -> z
.2(y, .2(i1(y), z)) -> z
.2(.2(x, y), z) -> .2(x, .2(y, z))
i1(.2(x, y)) -> .2(i1(y), i1(x))

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.