Termination w.r.t. Q proof of TRS/D3317.trs

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:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(x))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

.¹(.(x, y), z) → .¹(y, z)
.¹(.(x, y), z) → .¹(x, .(y, z))

The TRS R consists of the following rules:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(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.

.¹(.(x, y), z) → .¹(y, z)
.¹(.(x, y), z) → .¹(x, .(y, z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
.¹(x1, x2) = .¹(x1)
.(x1, x2) = .(x1, x2)
1 = 1
i(x1) = i(x1)

Recursive path order with status [2].
Precedence:

.₂ > .^1₁
1 > .^1₁
i₁ > .^1₁

Status:

i₁: multiset
.^1₁: multiset
1: multiset
.₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

I(.(x, y)) → I(y)
I(.(x, y)) → I(x)

The TRS R consists of the following rules:

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(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.

I(.(x, y)) → I(y)
I(.(x, y)) → I(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
I(x1) = I(x1)
.(x1, x2) = .(x1, x2)

Recursive path order with status [2].
Precedence:

.₂ > I₁

Status:

I₁: multiset
.₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

.(1, x) → x
.(x, 1) → x
.(i(x), x) → 1
.(x, i(x)) → 1
i(1) → 1
i(i(x)) → x
.(i(y), .(y, z)) → z
.(y, .(i(y), z)) → z
.(.(x, y), z) → .(x, .(y, z))
i(.(x, y)) → .(i(y), i(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.