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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


QUOT3(s1(x), s1(y), z) -> QUOT3(x, y, z)
The remaining pairs can at least by weakly be oriented.

QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
DIV2(x, y) -> QUOT3(x, y, y)
Used ordering: Combined order from the following AFS and order.
QUOT3(x1, x2, x3)  =  x1
0  =  0
s1(x1)  =  s1(x1)
DIV2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
0 > s1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


QUOT3(x, 0, s1(z)) -> DIV2(x, s1(z))
The remaining pairs can at least by weakly be oriented.

DIV2(x, y) -> QUOT3(x, y, y)
Used ordering: Combined order from the following AFS and order.
QUOT3(x1, x2, x3)  =  QUOT2(x1, x2)
0  =  0
s1(x1)  =  s
DIV2(x1, x2)  =  DIV2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[QUOT2, 0, DIV2] > s


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
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 [13,14,18] contains 0 SCCs with 1 less node.