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.

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 contains 2 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS2(x, s1(y)) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, s1(z))
Used argument filtering: TIMES2(x1, x2)  =  x2
plus2(x1, x2)  =  plus2(x1, x2)
s1(x1)  =  x1
times2(x1, x2)  =  times
0  =  0
Used ordering: Quasi Precedence: plus_2 > [times, 0]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TIMES2(x, plus2(y, s1(z))) -> TIMES2(x, plus2(y, times2(s1(z), 0)))
TIMES2(x, s1(y)) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2)  =  x2
plus2(x1, x2)  =  plus2(x1, x2)
s1(x1)  =  s1(x1)
times2(x1, x2)  =  times
0  =  0
Used ordering: Quasi Precedence: plus_2 > [s_1, times, 0]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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.