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, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(x, s1(y)) -> TIMES2(x, y)
TIMES2(x, s1(y)) -> PLUS2(times2(x, y), x)
PLUS2(x, s1(y)) -> PLUS2(x, y)

The TRS R consists of the following rules:

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

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:

PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(x, s1(y)) -> TIMES2(x, y)
TIMES2(x, s1(y)) -> PLUS2(times2(x, y), x)
PLUS2(x, s1(y)) -> PLUS2(x, y)

The TRS R consists of the following rules:

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

PLUS2(s1(x), y) -> PLUS2(x, y)
PLUS2(x, s1(y)) -> PLUS2(x, y)

The TRS R consists of the following rules:

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

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.


PLUS2(s1(x), y) -> PLUS2(x, y)
PLUS2(x, s1(y)) -> PLUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = 2·x1 + x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

TIMES2(x, s1(y)) -> TIMES2(x, y)

The TRS R consists of the following rules:

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

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, s1(y)) -> TIMES2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TIMES2(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(0, x) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
plus2(s1(x), y) -> s1(plus2(x, y))

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.