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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ODD(s(s(x))) → ODD(x)
-1(s(x), s(y)) → -1(x, y)
HALF(s(s(x))) → HALF(x)
F(x, s(y), z) → F(x, y, *(x, z))
F(x, s(y), z) → *1(x, x)
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
POW(x, y) → F(x, y, s(0))
F(x, s(y), z) → ODD(s(y))
F(x, s(y), z) → IF(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
*1(x, s(y)) → *1(x, y)
F(x, s(y), z) → HALF(s(y))
F(x, s(y), z) → *1(x, z)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(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:

ODD(s(s(x))) → ODD(x)
-1(s(x), s(y)) → -1(x, y)
HALF(s(s(x))) → HALF(x)
F(x, s(y), z) → F(x, y, *(x, z))
F(x, s(y), z) → *1(x, x)
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
POW(x, y) → F(x, y, s(0))
F(x, s(y), z) → ODD(s(y))
F(x, s(y), z) → IF(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
*1(x, s(y)) → *1(x, y)
F(x, s(y), z) → HALF(s(y))
F(x, s(y), z) → *1(x, z)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(HALF(x1)) = (1/2)x_1   
POL(s(x1)) = 1/4 + (2)x_1   
The value of delta used in the strict ordering is 3/8.
The following usable rules [17] 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(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:

ODD(s(s(x))) → ODD(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ODD(s(s(x))) → ODD(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ODD(x1)) = (1/2)x_1   
POL(s(x1)) = 1/4 + (2)x_1   
The value of delta used in the strict ordering is 3/8.
The following usable rules [17] 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(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:

*1(x, s(y)) → *1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


*1(x, s(y)) → *1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(*1(x1, x2)) = (1/4)x_2   
POL(s(x1)) = 1/4 + (2)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(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:

F(x, s(y), z) → F(x, y, *(x, z))
F(x, s(y), z) → F(*(x, x), half(s(y)), z)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, s(y), z) → F(x, y, *(x, z))
The remaining pairs can at least be oriented weakly.

F(x, s(y), z) → F(*(x, x), half(s(y)), z)
Used ordering: Polynomial interpretation [25,35]:

POL(half(x1)) = x_1   
POL(*(x1, x2)) = 0   
POL(s(x1)) = 1/4 + x_1   
POL(F(x1, x2, x3)) = (1/4)x_2   
POL(0) = 0   
POL(+(x1, x2)) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

half(s(0)) → 0
half(0) → 0
half(s(s(x))) → s(half(x))



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

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

F(x, s(y), z) → F(*(x, x), half(s(y)), z)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, s(y), z) → F(*(x, x), half(s(y)), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(half(x1)) = (1/4)x_1   
POL(*(x1, x2)) = 0   
POL(s(x1)) = 1/4 + (4)x_1   
POL(F(x1, x2, x3)) = (4)x_2   
POL(0) = 0   
POL(+(x1, x2)) = 0   
The value of delta used in the strict ordering is 3/4.
The following usable rules [17] were oriented:

half(s(0)) → 0
half(0) → 0
half(s(s(x))) → s(half(x))



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

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(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:

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


-1(s(x), s(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(-1(x1, x2)) = x_2   
POL(s(x1)) = 1 + x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
if(true, x, y) → x
if(false, x, y) → y
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if(true, x, y) → true
if(false, x, y) → false
pow(x, y) → f(x, y, s(0))
f(x, 0, z) → z
f(x, s(y), z) → if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(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.