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 strictly oriented and are deleted.


PLUS2(x, s1(y)) -> PLUS2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS2(x1, x2)  =  PLUS1(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[PLUS1, s1]


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 strictly oriented and are deleted.


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

TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)
Used ordering: Combined order from the following AFS and order.
TIMES2(x1, x2)  =  x2
plus2(x1, x2)  =  plus2(x1, x2)
s1(x1)  =  x1
times2(x1, x2)  =  x2
0  =  0

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


The following usable rules [14] were oriented:

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



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

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

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 strictly oriented and are deleted.


TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)
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)
s1(x1)  =  s1(x1)
times2(x1, x2)  =  times
0  =  0

Lexicographic Path Order [19].
Precedence:
plus2 > TIMES1 > [s1, times, 0]


The following usable rules [14] were oriented:

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



↳ 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.