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 Order [17,21] with Interpretation:

POL( 1 ) = 2


POL( 0 ) = 1


POL( TIMES2(x1, x2) ) = x1 + x2


POL( times2(x1, x2) ) = max{0, 2x2 - 1}


POL( plus2(x1, x2) ) = x1 + x2



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.