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:

admit2(x, nil) -> nil
admit2(x, .2(u, .2(v, .2(w, z)))) -> cond2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))
cond2(true, y) -> y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

admit2(x, nil) -> nil
admit2(x, .2(u, .2(v, .2(w, z)))) -> cond2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))
cond2(true, y) -> y

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> ADMIT2(carry3(x, u, v), z)
ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> COND2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))

The TRS R consists of the following rules:

admit2(x, nil) -> nil
admit2(x, .2(u, .2(v, .2(w, z)))) -> cond2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))
cond2(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:

ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> ADMIT2(carry3(x, u, v), z)
ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> COND2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))

The TRS R consists of the following rules:

admit2(x, nil) -> nil
admit2(x, .2(u, .2(v, .2(w, z)))) -> cond2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))
cond2(true, y) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> ADMIT2(carry3(x, u, v), z)

The TRS R consists of the following rules:

admit2(x, nil) -> nil
admit2(x, .2(u, .2(v, .2(w, z)))) -> cond2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))
cond2(true, y) -> y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> ADMIT2(carry3(x, u, v), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 2 + x1 + 2·x2   
POL(ADMIT2(x1, x2)) = x1 + 2·x2   
POL(carry3(x1, x2, x3)) = 1 + x2 + 2·x3   
POL(w) = 0   

The following usable rules [14] 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:

admit2(x, nil) -> nil
admit2(x, .2(u, .2(v, .2(w, z)))) -> cond2(=2(sum3(x, u, v), w), .2(u, .2(v, .2(w, admit2(carry3(x, u, v), z)))))
cond2(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.