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)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 3 less nodes.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


TIMES2(x, plus2(y, 1)) -> TIMES2(x, plus2(y, times2(1, 0)))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TIMES2(x1, x2)  =  TIMES1(x2)
plus2(x1, x2)  =  plus2(x1, x2)
1  =  1
times2(x1, x2)  =  x2
0  =  0

Lexicographic Path Order [19].
Precedence:
[TIMES1, 1] > 0 > plus2


The following usable rules [14] were oriented:

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



↳ 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:

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.