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:

-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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


HALF1(s1(s1(x))) -> HALF1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( HALF1(x1) ) = max{0, x1 - 3}


POL( s1(x1) ) = x1 + 2



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(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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ODD1(s1(s1(x))) -> ODD1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ODD1(x1) ) = max{0, x1 - 3}


POL( s1(x1) ) = x1 + 2



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(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
            ↳ 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(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.
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 Order [17,21] with Interpretation:

POL( *12(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



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(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
            ↳ QDPOrderProof
          ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F3(x1, ..., x3) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3


POL( half1(x1) ) = max{0, x1 - 3}


POL( 0 ) = 0



The following usable rules [14] were oriented:

half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
half1(0) -> 0



↳ 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(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
          ↳ QDP
            ↳ QDPOrderProof

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.
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 Order [17,21] with Interpretation:

POL( -12(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



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