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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

admit2(x0, nil)
admit2(x0, .2(x1, .2(x2, .2(w, x3))))
cond2(true, x0)


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

The set Q consists of the following terms:

admit2(x0, nil)
admit2(x0, .2(x1, .2(x2, .2(w, x3))))
cond2(true, x0)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

admit2(x0, nil)
admit2(x0, .2(x1, .2(x2, .2(w, x3))))
cond2(true, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

The set Q consists of the following terms:

admit2(x0, nil)
admit2(x0, .2(x1, .2(x2, .2(w, x3))))
cond2(true, x0)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> ADMIT2(carry3(x, u, v), z)
Used argument filtering: ADMIT2(x1, x2)  =  x2
.2(x1, x2)  =  .1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ 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

The set Q consists of the following terms:

admit2(x0, nil)
admit2(x0, .2(x1, .2(x2, .2(w, x3))))
cond2(true, x0)

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