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:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

Q is empty.

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

DIV2(div2(x, y), z) -> TIMES2(y, z)
PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(s1(x), y) -> PLUS2(y, times2(x, y))
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
TIMES2(s1(x), y) -> TIMES2(x, y)
DIV2(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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:

DIV2(div2(x, y), z) -> TIMES2(y, z)
PLUS2(s1(x), y) -> PLUS2(x, y)
TIMES2(s1(x), y) -> PLUS2(y, times2(x, y))
QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
TIMES2(s1(x), y) -> TIMES2(x, y)
DIV2(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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


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

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

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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
          ↳ QDP

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

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

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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(s1(x), y) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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
          ↳ QDP
            ↳ QDPAfsSolverProof

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

QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
DIV2(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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.

QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
Used argument filtering: QUOT3(x1, x2, x3)  =  x1
DIV2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
div2(x1, x2)  =  x1
times2(x1, x2)  =  times2(x1, x2)
0  =  0
plus2(x1, x2)  =  plus2(x1, x2)
Used ordering: Quasi Precedence: times_2 > plus_2 > s_1 


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

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

QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
DIV2(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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.

DIV2(div2(x, y), z) -> DIV2(x, times2(y, z))
Used argument filtering: QUOT3(x1, x2, x3)  =  x1
DIV2(x1, x2)  =  x1
div2(x1, x2)  =  div1(x1)
times2(x1, x2)  =  times2(x1, x2)
0  =  0
s1(x1)  =  s1(x1)
plus2(x1, x2)  =  plus2(x1, x2)
Used ordering: Quasi Precedence: times_2 > plus_2 > s_1 


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

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

QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
DIV2(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

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.

QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
Used argument filtering: QUOT3(x1, x2, x3)  =  x2
0  =  0
DIV2(x1, x2)  =  x2
s1(x1)  =  s
Used ordering: Quasi Precedence: 0 > s 


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

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

DIV2(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(s1(0), y) -> y
times2(s1(x), y) -> plus2(y, times2(x, y))
div2(0, y) -> 0
div2(x, y) -> quot3(x, y, y)
quot3(0, s1(y), z) -> 0
quot3(s1(x), s1(y), z) -> quot3(x, y, z)
quot3(x, 0, s1(z)) -> s1(div2(x, s1(z)))
div2(div2(x, y), z) -> div2(x, times2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.