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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • .1(.(x, y), z) → .1(y, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

  • .1(.(x, y), z) → .1(x, .(y, z)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • I(.(x, y)) → I(x) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • I(.(x, y)) → I(y) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE