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:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

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:

ADMIT(x, .(u, .(v, .(w, z)))) → COND(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)

The TRS R consists of the following rules:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

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 1 SCC with 1 less node.

(4) Obligation:

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

ADMIT(x, .(u, .(v, .(w, z)))) → ADMIT(carry(x, u, v), z)

The TRS R consists of the following rules:

admit(x, nil) → nil
admit(x, .(u, .(v, .(w, z)))) → cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) → y

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

(5) 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:
w  =  w
.(x1, x2)  =  .(x2)
carry(x1, x2, x3)  =  carry

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

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

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

(6) TRUE