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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔)
        ↳11 TRUE
    ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE

(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)
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(x0, x2)

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

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(.(x1, x2)) = 1 + x1 + x2   
POL(.1(x1, x2)) = x1 + x2   
POL(1) = 1   
POL(i(x1)) = 1   

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:
The TRS P consists of the following rules:

.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.

(8) 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(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 [1,1,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(.(x1, x2)) = 1 + x1   
POL(.1(x1, x2)) = 1   
POL(1) = 0   
POL(i(x1)) = 1   

The following usable rules [FROCOS05] were oriented: none

(9) 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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.

(13) 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:
Polynomial interpretation [POLO]:

POL(.(x1, x2)) = 1 + x1 + x2   
POL(I(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(14) 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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE