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:

-(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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

F(x, s(y), z) → ODD(s(y))
POW(x, y) → F(x, y, s(0))
F(x, s(y), z) → *1(x, x)
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
*1(x, s(y)) → *1(x, y)
-1(s(x), s(y)) → -1(x, y)
F(x, s(y), z) → F(x, y, *(x, z))
F(x, s(y), z) → HALF(s(y))
F(x, s(y), z) → *1(x, z)
ODD(s(s(x))) → ODD(x)
F(x, s(y), z) → IF(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
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 deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

*1(x, s(y)) → *1(x, y)
F(x, s(y), z) → *1(x, x)
-1(s(x), s(y)) → -1(x, y)
F(x, s(y), z) → HALF(s(y))
ODD(s(s(x))) → ODD(x)
F(x, s(y), z) → IF(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
F(x, s(y), z) → ODD(s(y))
POW(x, y) → F(x, y, s(0))
F(x, s(y), z) → F(*(x, x), half(s(y)), z)
F(x, s(y), z) → F(x, y, *(x, z))
F(x, s(y), z) → *1(x, z)
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.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
HALF(x1)  =  HALF(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
HALF1: [1]
s1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
ODD(x1)  =  ODD(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
s1: multiset
ODD1: [1]

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > *^12

Status:
*^12: [1,2]
s1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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, x), half(s(y)), z)
F(x, s(y), z) → F(x, y, *(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.
We use the reduction pair processor [13].


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: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x2)
s(x1)  =  s(x1)
*(x1, x2)  =  *
half(x1)  =  x1
0  =  0
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [19].
Precedence:
F1 > +2
s1 > +2
* > +2
0 > +2

Status:
+2: multiset
s1: [1]
0: multiset
F1: [1]
*: []

The following usable rules [14] were oriented:

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



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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
-1(x1, x2)  =  -1(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > -^11

Status:
-^11: [1]
s1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.