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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(s1(z), 0)
TIMES2(x, s1(y)) -> TIMES2(x, y)
TIMES2(x, s1(y)) -> PLUS2(times2(x, y), x)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(y, times2(s1(z), 0))
PLUS2(x, s1(y)) -> PLUS2(x, y)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))

The TRS R consists of the following rules:

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

TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(s1(z), 0)
TIMES2(x, s1(y)) -> TIMES2(x, y)
TIMES2(x, s1(y)) -> PLUS2(times2(x, y), x)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(y, times2(s1(z), 0))
PLUS2(x, s1(y)) -> PLUS2(x, y)
TIMES2(x, plus2(y, s1(z))) -> PLUS2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))

The TRS R consists of the following rules:

times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(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 4 less nodes.

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

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

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

The TRS R consists of the following rules:

times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(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(x, s1(y)) -> PLUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( PLUS2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



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, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(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, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)

The TRS R consists of the following rules:

times2(x, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(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, plus2(y, s1(z))) -> TIMES2(x, s1(z))
TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
The remaining pairs can at least be oriented weakly.

TIMES2(x, s1(y)) -> TIMES2(x, y)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( TIMES2(x1, x2) ) = max{0, x2 - 1}


POL( plus2(x1, x2) ) = x1 + x2 + 1


POL( s1(x1) ) = x1 + 1


POL( times2(x1, x2) ) = 0


POL( 0 ) = 0



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
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, plus2(y, s1(z))) -> plus2(times2(x, plus2(y, times2(s1(z), 0))), times2(x, s1(z)))
times2(x, 0) -> 0
times2(x, s1(y)) -> plus2(times2(x, y), x)
plus2(x, 0) -> x
plus2(x, s1(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 Order [17,21] with Interpretation:

POL( TIMES2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

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