Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ADMIT(x1, x2)) = (4)x_2   
POL(carry(x1, x2, x3)) = 3 + (2)x_1 + (4)x_2 + (4)x_3   
POL(w) = 2   
POL(.(x1, x2)) = x_1 + (4)x_2   
The value of delta used in the strict ordering is 128.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.