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
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

HALF(s(s(x))) → HALF(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ODD(s(s(x))) → ODD(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

*1(x, s(y)) → *1(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ 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]:

POL(*(x1, x2)) = 0   
POL(+(x1, x2)) = 0   
POL(0) = 1   
POL(F(x1, x2, x3)) = x2   
POL(half(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

half(0) → 0
half(s(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: Matrix interpretation [3]:
Non-tuple symbols:
M( half(x1) ) =
/0\
\0/
+
/01\
\01/
·x1

M( *(x1, x2) ) =
/0\
\0/
+
/11\
\00/
·x1+
/01\
\01/
·x2

M( s(x1) ) =
/1\
\0/
+
/11\
\11/
·x1

M( 0 ) =
/0\
\0/

M( +(x1, x2) ) =
/0\
\0/
+
/01\
\01/
·x1+
/00\
\00/
·x2

Tuple symbols:
M( F(x1, ..., x3) ) = 0+
[0,1]
·x1+
[1,0]
·x2+
[0,0]
·x3


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

*(x, 0) → 0
*(x, s(y)) → +(*(x, y), x)
half(0) → 0
half(s(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
            ↳ UsableRulesProof

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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof

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

-1(s(x), s(y)) → -1(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: