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)


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

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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ADMIT2(x, .2(u, .2(v, .2(w, z)))) -> ADMIT2(carry3(x, u, v), z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADMIT2(x1, x2)  =  x2
.2(x1, x2)  =  .1(x2)
w  =  w
carry3(x1, x2, x3)  =  carry3(x1, x2, x3)

Lexicographic Path Order [19].
Precedence:
.1 > carry3
w > carry3

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

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.