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:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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:

DIV_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
IF_ACTIVE3(true, x, y) -> MARK1(x)
GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
MARK1(ge2(x, y)) -> GE_ACTIVE2(x, y)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(div2(x, y)) -> MARK1(x)
MARK1(minus2(x, y)) -> MINUS_ACTIVE2(x, y)
MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
IF_ACTIVE3(false, x, y) -> MARK1(y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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:

DIV_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
IF_ACTIVE3(true, x, y) -> MARK1(x)
GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
MARK1(ge2(x, y)) -> GE_ACTIVE2(x, y)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(div2(x, y)) -> MARK1(x)
MARK1(minus2(x, y)) -> MINUS_ACTIVE2(x, y)
MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
IF_ACTIVE3(false, x, y) -> MARK1(y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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 3 SCCs with 3 less nodes.

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

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

GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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.


GE_ACTIVE2(s1(x), s1(y)) -> GE_ACTIVE2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
GE_ACTIVE2(x1, x2)  =  GE_ACTIVE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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
          ↳ QDP

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

MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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.


MINUS_ACTIVE2(s1(x), s1(y)) -> MINUS_ACTIVE2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS_ACTIVE2(x1, x2)  =  MINUS_ACTIVE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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
          ↳ QDP
            ↳ QDPOrderProof

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

MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
IF_ACTIVE3(true, x, y) -> MARK1(x)
MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(div2(x, y)) -> MARK1(x)
IF_ACTIVE3(false, x, y) -> MARK1(y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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.


MARK1(if3(x, y, z)) -> MARK1(x)
MARK1(s1(x)) -> MARK1(x)
IF_ACTIVE3(true, x, y) -> MARK1(x)
MARK1(div2(x, y)) -> MARK1(x)
IF_ACTIVE3(false, x, y) -> MARK1(y)
DIV_ACTIVE2(s1(x), s1(y)) -> IF_ACTIVE3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
The remaining pairs can at least by weakly be oriented.

MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  x1
if3(x1, x2, x3)  =  if3(x1, x2, x3)
s1(x1)  =  s1(x1)
IF_ACTIVE3(x1, x2, x3)  =  IF_ACTIVE3(x1, x2, x3)
true  =  true
div2(x1, x2)  =  div2(x1, x2)
DIV_ACTIVE2(x1, x2)  =  DIV_ACTIVE2(x1, x2)
mark1(x1)  =  x1
false  =  false
ge_active2(x1, x2)  =  ge_active2(x1, x2)
minus2(x1, x2)  =  x1
0  =  0
minus_active2(x1, x2)  =  x1
if_active3(x1, x2, x3)  =  if_active3(x1, x2, x3)
div_active2(x1, x2)  =  div_active2(x1, x2)
ge2(x1, x2)  =  ge2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[div2, DIVACTIVE2, 0, divactive2] > s1 > [if3, IFACTIVE3, ifactive3]
[div2, DIVACTIVE2, 0, divactive2] > s1 > [true, false, geactive2, ge2]


The following usable rules [14] were oriented:

ge_active2(x, 0) -> true
ge_active2(0, s1(y)) -> false
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
ge_active2(x, y) -> ge2(x, y)
mark1(0) -> 0
mark1(s1(x)) -> s1(mark1(x))
mark1(minus2(x, y)) -> minus_active2(x, y)
mark1(ge2(x, y)) -> ge_active2(x, y)
if_active3(true, x, y) -> mark1(x)
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
if_active3(false, x, y) -> mark1(y)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
div_active2(x, y) -> div2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
minus_active2(0, y) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
minus_active2(x, y) -> minus2(x, y)



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

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

MARK1(div2(x, y)) -> DIV_ACTIVE2(mark1(x), y)
MARK1(if3(x, y, z)) -> IF_ACTIVE3(mark1(x), y, z)

The TRS R consists of the following rules:

minus_active2(0, y) -> 0
mark1(0) -> 0
minus_active2(s1(x), s1(y)) -> minus_active2(x, y)
mark1(s1(x)) -> s1(mark1(x))
ge_active2(x, 0) -> true
mark1(minus2(x, y)) -> minus_active2(x, y)
ge_active2(0, s1(y)) -> false
mark1(ge2(x, y)) -> ge_active2(x, y)
ge_active2(s1(x), s1(y)) -> ge_active2(x, y)
mark1(div2(x, y)) -> div_active2(mark1(x), y)
div_active2(0, s1(y)) -> 0
mark1(if3(x, y, z)) -> if_active3(mark1(x), y, z)
div_active2(s1(x), s1(y)) -> if_active3(ge_active2(x, y), s1(div2(minus2(x, y), s1(y))), 0)
if_active3(true, x, y) -> mark1(x)
minus_active2(x, y) -> minus2(x, y)
if_active3(false, x, y) -> mark1(y)
ge_active2(x, y) -> ge2(x, y)
if_active3(x, y, z) -> if3(x, y, z)
div_active2(x, y) -> div2(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 0 SCCs with 2 less nodes.