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:

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

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:

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

The set Q consists of the following terms:

times2(x0, plus2(x1, 1))
times2(x0, 1)
plus2(x0, 0)
times2(x0, 0)


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

TIMES2(x, plus2(y, 1)) -> TIMES2(x, plus2(y, times2(1, 0)))
TIMES2(x, plus2(y, 1)) -> TIMES2(1, 0)
TIMES2(x, plus2(y, 1)) -> PLUS2(times2(x, plus2(y, times2(1, 0))), x)
TIMES2(x, plus2(y, 1)) -> PLUS2(y, times2(1, 0))

The TRS R consists of the following rules:

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

The set Q consists of the following terms:

times2(x0, plus2(x1, 1))
times2(x0, 1)
plus2(x0, 0)
times2(x0, 0)

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:

TIMES2(x, plus2(y, 1)) -> TIMES2(x, plus2(y, times2(1, 0)))
TIMES2(x, plus2(y, 1)) -> TIMES2(1, 0)
TIMES2(x, plus2(y, 1)) -> PLUS2(times2(x, plus2(y, times2(1, 0))), x)
TIMES2(x, plus2(y, 1)) -> PLUS2(y, times2(1, 0))

The TRS R consists of the following rules:

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

The set Q consists of the following terms:

times2(x0, plus2(x1, 1))
times2(x0, 1)
plus2(x0, 0)
times2(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

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

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

TIMES2(x, plus2(y, 1)) -> TIMES2(x, plus2(y, times2(1, 0)))

The TRS R consists of the following rules:

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

The set Q consists of the following terms:

times2(x0, plus2(x1, 1))
times2(x0, 1)
plus2(x0, 0)
times2(x0, 0)

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.

TIMES2(x, plus2(y, 1)) -> TIMES2(x, plus2(y, times2(1, 0)))
Used argument filtering: TIMES2(x1, x2)  =  x2
plus2(x1, x2)  =  plus2(x1, x2)
1  =  1
times2(x1, x2)  =  times
0  =  0
Used ordering: Precedence:
1 > times > 0



↳ 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:

times2(x, plus2(y, 1)) -> plus2(times2(x, plus2(y, times2(1, 0))), x)
times2(x, 1) -> x
plus2(x, 0) -> x
times2(x, 0) -> 0

The set Q consists of the following terms:

times2(x0, plus2(x1, 1))
times2(x0, 1)
plus2(x0, 0)
times2(x0, 0)

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