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))


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
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

BIN2(s1(x), s1(y)) -> BIN2(x, y)
Used argument filtering: BIN2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

BIN2(s1(x), s1(y)) -> BIN2(x, s1(y))
Used argument filtering: BIN2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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