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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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

The set Q consists of the following terms:

admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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

The set Q consists of the following terms:

admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) 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

The set Q consists of the following terms:

admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADMIT(x1, x2)  =  ADMIT(x1, x2)
.(x1, x2)  =  .(x2)
w  =  w
carry(x1, x2, x3)  =  carry
admit(x1, x2)  =  x2
nil  =  nil
cond(x1, x2)  =  x2
=(x1, x2)  =  =(x1, x2)
sum(x1, x2, x3)  =  sum(x1)
true  =  true

Recursive Path Order [RPO].
Precedence:

w > ADMIT₂ > carry > =₂
w > .₁ > carry > =₂
w > sum₁ > =₂
nil > =₂
true > =₂

The following usable rules [FROCOS05] were oriented:

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

(8) Obligation:

Q DP problem:
P is empty.
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

The set Q consists of the following terms:

admit(x0, nil)
admit(x0, .(x1, .(x2, .(w, x3))))
cond(true, x0)

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE