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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))

Q is empty.

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

F3(x, s1(y), z) -> ODD1(s1(y))
POW2(x, y) -> F3(x, y, s1(0))
F3(x, s1(y), z) -> *12(x, x)
F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F3(x, s1(y), z) -> F3(x, y, *2(x, z))
F3(x, s1(y), z) -> HALF1(s1(y))
F3(x, s1(y), z) -> *12(x, z)
ODD1(s1(s1(x))) -> ODD1(x)
F3(x, s1(y), z) -> IF3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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:

F3(x, s1(y), z) -> ODD1(s1(y))
POW2(x, y) -> F3(x, y, s1(0))
F3(x, s1(y), z) -> *12(x, x)
F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F3(x, s1(y), z) -> F3(x, y, *2(x, z))
F3(x, s1(y), z) -> HALF1(s1(y))
F3(x, s1(y), z) -> *12(x, z)
ODD1(s1(s1(x))) -> ODD1(x)
F3(x, s1(y), z) -> IF3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))
HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 6 less nodes.

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

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

HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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.

HALF1(s1(s1(x))) -> HALF1(x)
Used argument filtering: HALF1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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
          ↳ QDP
          ↳ QDP

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

ODD1(s1(s1(x))) -> ODD1(x)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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.

ODD1(s1(s1(x))) -> ODD1(x)
Used argument filtering: ODD1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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
          ↳ 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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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.

*12(x, s1(y)) -> *12(x, y)
Used argument filtering: *12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)
F3(x, s1(y), z) -> F3(x, y, *2(x, z))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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.

F3(x, s1(y), z) -> F3(x, y, *2(x, z))
Used argument filtering: F3(x1, x2, x3)  =  x2
s1(x1)  =  s1(x1)
half1(x1)  =  x1
*2(x1, x2)  =  *
0  =  0
+2(x1, x2)  =  +
Used ordering: Quasi Precedence: * > 0 * > +


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

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

F3(x, s1(y), z) -> F3(*2(x, x), half1(s1(y)), z)

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(y)), z))

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

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

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(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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.

-12(s1(x), s1(y)) -> -12(x, y)
Used argument filtering: -12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
if3(true, x, y) -> x
if3(false, x, y) -> y
odd1(0) -> false
odd1(s1(0)) -> true
odd1(s1(s1(x))) -> odd1(x)
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
if3(true, x, y) -> true
if3(false, x, y) -> false
pow2(x, y) -> f3(x, y, s1(0))
f3(x, 0, z) -> z
f3(x, s1(y), z) -> if3(odd1(s1(y)), f3(x, y, *2(x, z)), f3(*2(x, x), half1(s1(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.