Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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:

IFFACT2(x, true) -> *12(x, fact1(-2(x, s1(0))))
*12(x, s1(y)) -> +12(*2(x, y), x)
+12(x, s1(y)) -> +12(x, y)
*12(x, s1(y)) -> *12(x, y)
FACT1(x) -> IFFACT2(x, ge2(x, s1(s1(0))))
-12(s1(x), s1(y)) -> -12(x, y)
IFFACT2(x, true) -> FACT1(-2(x, s1(0)))
IFFACT2(x, true) -> -12(x, s1(0))
FACT1(x) -> GE2(x, s1(s1(0)))
GE2(s1(x), s1(y)) -> GE2(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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:

IFFACT2(x, true) -> *12(x, fact1(-2(x, s1(0))))
*12(x, s1(y)) -> +12(*2(x, y), x)
+12(x, s1(y)) -> +12(x, y)
*12(x, s1(y)) -> *12(x, y)
FACT1(x) -> IFFACT2(x, ge2(x, s1(s1(0))))
-12(s1(x), s1(y)) -> -12(x, y)
IFFACT2(x, true) -> FACT1(-2(x, s1(0)))
IFFACT2(x, true) -> -12(x, s1(0))
FACT1(x) -> GE2(x, s1(s1(0)))
GE2(s1(x), s1(y)) -> GE2(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 4 less nodes.

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

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

-12(s1(x), s1(y)) -> -12(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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 oriented strictly and are deleted.


-12(s1(x), s1(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-12(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



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

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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
          ↳ QDP
          ↳ QDP

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

GE2(s1(x), s1(y)) -> GE2(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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 oriented strictly and are deleted.


GE2(s1(x), s1(y)) -> GE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(GE2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



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

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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
          ↳ QDP
          ↳ QDP

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

+12(x, s1(y)) -> +12(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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 oriented strictly and are deleted.


+12(x, s1(y)) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+12(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

*12(x, s1(y)) -> *12(x, y)

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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 oriented strictly and are deleted.


*12(x, s1(y)) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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
          ↳ QDP
          ↳ QDP

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

FACT1(x) -> IFFACT2(x, ge2(x, s1(s1(0))))
IFFACT2(x, true) -> FACT1(-2(x, s1(0)))

The TRS R consists of the following rules:

+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
ge2(x, 0) -> true
ge2(0, s1(y)) -> false
ge2(s1(x), s1(y)) -> ge2(x, y)
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
fact1(x) -> iffact2(x, ge2(x, s1(s1(0))))
iffact2(x, true) -> *2(x, fact1(-2(x, s1(0))))
iffact2(x, false) -> s1(0)

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