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
  ↳ 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

Q is empty.

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

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

↳ 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

Q is empty.
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
  ↳ 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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TIMES2(x, plus2(y, 1)) -> TIMES2(x, plus2(y, times2(1, 0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(1) = 2   
POL(TIMES2(x1, x2)) = x1·x2 + 3·x2   
POL(plus2(x1, x2)) = x1 + x1·x2 + x2   
POL(times2(x1, x2)) = 1   

The following usable rules [14] were oriented:

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



↳ 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

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