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:

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)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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)))

Q is empty.

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

The TRS R consists of the following rules:

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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
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(x, y) -> QUOT3(x, y, y)

The TRS R consists of the following rules:

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)))

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)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

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)))

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
      ↳ 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:

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)))

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.