Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
.¹(x0, x1, x2)  =  .¹(x1)

Tags:
.¹ has argument tags [0,3,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
.¹(x1, x2)  =  .¹(x2)
.(x1, x2)  =  .(x1, x2)
1  =  1
i(x1)  =  i

Lexicographic path order with status [LPO].
Quasi-Precedence:

[.^1₁, .₂] > 1

Status:

.^1₁: [1]
.₂: [1,2]
1: []
i: []

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

I(.(x, y)) → I(x)
I(.(x, y)) → I(y)

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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

I(.(x, y)) → I(x)
I(.(x, y)) → I(y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
I(x0, x1)  =  I(x1)

Tags:
I has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
I(x1)  =  I
.(x1, x2)  =  .(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

.₂ > I

Status:

I: []
.₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE