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:

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, 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:

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, y))

The set Q consists of the following terms:

bin2(x0, 0)
bin2(0, s1(x0))
bin2(s1(x0), s1(x1))


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

BIN2(s1(x), s1(y)) -> BIN2(x, y)
BIN2(s1(x), s1(y)) -> BIN2(x, s1(y))

The TRS R consists of the following rules:

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, y))

The set Q consists of the following terms:

bin2(x0, 0)
bin2(0, s1(x0))
bin2(s1(x0), s1(x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

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

BIN2(s1(x), s1(y)) -> BIN2(x, y)
BIN2(s1(x), s1(y)) -> BIN2(x, s1(y))

The TRS R consists of the following rules:

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, y))

The set Q consists of the following terms:

bin2(x0, 0)
bin2(0, s1(x0))
bin2(s1(x0), s1(x1))

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.


BIN2(s1(x), s1(y)) -> BIN2(x, y)
The remaining pairs can at least by weakly be oriented.

BIN2(s1(x), s1(y)) -> BIN2(x, s1(y))
Used ordering: Combined order from the following AFS and order.
BIN2(x1, x2)  =  BIN1(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > BIN1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

BIN2(s1(x), s1(y)) -> BIN2(x, s1(y))

The TRS R consists of the following rules:

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, y))

The set Q consists of the following terms:

bin2(x0, 0)
bin2(0, s1(x0))
bin2(s1(x0), s1(x1))

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.


BIN2(s1(x), s1(y)) -> BIN2(x, s1(y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
BIN2(x1, x2)  =  BIN1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[BIN1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

bin2(x, 0) -> s1(0)
bin2(0, s1(y)) -> 0
bin2(s1(x), s1(y)) -> +2(bin2(x, s1(y)), bin2(x, y))

The set Q consists of the following terms:

bin2(x0, 0)
bin2(0, s1(x0))
bin2(s1(x0), s1(x1))

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