Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
-(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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
*'(x, s(y)) -> *'(x, y)
ODD(s(s(x))) -> ODD(x)
HALF(s(s(x))) -> HALF(x)
POW(x, y) -> F(x, y, s(0))
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))
F(x, s(y), z) -> F(x, y, *(x, z))
F(x, s(y), z) -> *'(x, z)
F(x, s(y), z) -> F(*(x, x), half(s(y)), z)
F(x, s(y), z) -> *'(x, x)
F(x, s(y), z) -> HALF(s(y))

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

-'(s(x), s(y)) -> -'(x, y)

one new Dependency Pair is created:

-'(s(s(x'')), s(s(y''))) -> -'(s(x''), s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

-'(s(s(x'')), s(s(y''))) -> -'(s(x''), s(y''))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

-'(s(s(x'')), s(s(y''))) -> -'(s(x''), s(y''))

one new Dependency Pair is created:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(s(s(x'''')), s(s(y'''')))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

-'(s(s(s(x''''))), s(s(s(y'''')))) -> -'(s(s(x'''')), s(s(y'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

*'(x, s(y)) -> *'(x, y)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

*'(x, s(y)) -> *'(x, y)

one new Dependency Pair is created:

*'(x'', s(s(y''))) -> *'(x'', s(y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

*'(x'', s(s(y''))) -> *'(x'', s(y''))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

*'(x'', s(s(y''))) -> *'(x'', s(y''))

one new Dependency Pair is created:

*'(x'''', s(s(s(y'''')))) -> *'(x'''', s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 10

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

*'(x'''', s(s(s(y'''')))) -> *'(x'''', s(s(y'''')))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(x'''', s(s(s(y'''')))) -> *'(x'''', s(s(y'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = 1 + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 9

             ↳FwdInst

...

               →DP Problem 11

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

ODD(s(s(x))) -> ODD(x)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

ODD(s(s(x))) -> ODD(x)

one new Dependency Pair is created:

ODD(s(s(s(s(x''))))) -> ODD(s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

ODD(s(s(s(s(x''))))) -> ODD(s(s(x'')))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

ODD(s(s(s(s(x''))))) -> ODD(s(s(x'')))

one new Dependency Pair is created:

ODD(s(s(s(s(s(s(x''''))))))) -> ODD(s(s(s(s(x'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳FwdInst

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

ODD(s(s(s(s(s(s(x''''))))))) -> ODD(s(s(s(s(x'''')))))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

ODD(s(s(s(s(s(s(x''''))))))) -> ODD(s(s(s(s(x'''')))))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ODD(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 12

             ↳FwdInst

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

Dependency Pair:

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

HALF(s(s(x))) -> HALF(x)

one new Dependency Pair is created:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳Forward Instantiation Transformation

       →DP Problem 5

         ↳Nar

Dependency Pair:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))

one new Dependency Pair is created:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳FwdInst

...

               →DP Problem 16

                 ↳Polynomial Ordering

       →DP Problem 5

         ↳Nar

Dependency Pair:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(HALF(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 15

             ↳FwdInst

...

               →DP Problem 17

                 ↳Dependency Graph

       →DP Problem 5

         ↳Nar

Dependency Pair:

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Narrowing Transformation

Dependency Pairs:

F(x, s(y), z) -> F(*(x, x), half(s(y)), z)
F(x, s(y), z) -> F(x, y, *(x, z))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

F(x, s(y), z) -> F(*(x, x), half(s(y)), z)

four new Dependency Pairs are created:

F(0, s(y), z) -> F(0, half(s(y)), z)
F(s(y''), s(y), z) -> F(+(*(s(y''), y''), s(y'')), half(s(y)), z)
F(x, s(0), z) -> F(*(x, x), 0, z)
F(x, s(s(x'')), z) -> F(*(x, x), s(half(x'')), z)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Narrowing Transformation

Dependency Pairs:

F(s(y''), s(y), z) -> F(+(*(s(y''), y''), s(y'')), half(s(y)), z)
F(x, s(s(x'')), z) -> F(*(x, x), s(half(x'')), z)
F(0, s(y), z) -> F(0, half(s(y)), z)
F(x, s(y), z) -> F(x, y, *(x, z))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

F(0, s(y), z) -> F(0, half(s(y)), z)

two new Dependency Pairs are created:

F(0, s(0), z) -> F(0, 0, z)
F(0, s(s(x')), z) -> F(0, s(half(x')), z)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 19

                 ↳Narrowing Transformation

Dependency Pairs:

F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(x, s(s(x'')), z) -> F(*(x, x), s(half(x'')), z)
F(x, s(y), z) -> F(x, y, *(x, z))
F(s(y''), s(y), z) -> F(+(*(s(y''), y''), s(y'')), half(s(y)), z)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

F(s(y''), s(y), z) -> F(+(*(s(y''), y''), s(y'')), half(s(y)), z)

four new Dependency Pairs are created:

F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(s(y''), s(0), z) -> F(+(*(s(y''), y''), s(y'')), 0, z)
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 20

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(x, s(s(x'')), z) -> F(*(x, x), s(half(x'')), z)
F(x, s(y), z) -> F(x, y, *(x, z))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(x, s(y), z) -> F(x, y, *(x, z))

six new Dependency Pairs are created:

F(x'', s(s(y'')), z'') -> F(x'', s(y''), *(x'', z''))
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 21

                 ↳Narrowing Transformation

Dependency Pairs:

F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(x'', s(s(y'')), z'') -> F(x'', s(y''), *(x'', z''))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(x, s(s(x'')), z) -> F(*(x, x), s(half(x'')), z)
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

F(x, s(s(x'')), z) -> F(*(x, x), s(half(x'')), z)

five new Dependency Pairs are created:

F(0, s(s(x'')), z) -> F(0, s(half(x'')), z)
F(s(y'), s(s(x'')), z) -> F(+(*(s(y'), y'), s(y')), s(half(x'')), z)
F(x, s(s(0)), z) -> F(*(x, x), s(0), z)
F(x, s(s(s(0))), z) -> F(*(x, x), s(0), z)
F(x, s(s(s(s(x''')))), z) -> F(*(x, x), s(s(half(x'''))), z)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 22

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(0, s(s(x'')), z) -> F(0, s(half(x'')), z)
F(s(y'), s(s(x'')), z) -> F(+(*(s(y'), y'), s(y')), s(half(x'')), z)
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(x, s(s(s(s(x''')))), z) -> F(*(x, x), s(s(half(x'''))), z)
F(x, s(s(s(0))), z) -> F(*(x, x), s(0), z)
F(x, s(s(0)), z) -> F(*(x, x), s(0), z)
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(x'', s(s(y'')), z'') -> F(x'', s(y''), *(x'', z''))
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(x'', s(s(y'')), z'') -> F(x'', s(y''), *(x'', z''))

12 new Dependency Pairs are created:

F(x'''', s(s(s(x'''''))), z''') -> F(x'''', s(s(x''''')), *(x'''', z'''))
F(s(x'''), s(s(y''')), z''') -> F(s(x'''), s(y'''), *(s(x'''), z'''))
F(s(s(y''''')), s(s(y''')), z''') -> F(s(s(y''''')), s(y'''), *(s(s(y''''')), z'''))
F(s(y''''), s(s(s(x''''))), z''') -> F(s(y''''), s(s(x'''')), *(s(y''''), z'''))
F(x'''', s(s(s(y''''))), z'''') -> F(x'''', s(s(y'''')), *(x'''', z''''))
F(x'''', s(s(s(s(x'''''')))), z'''') -> F(x'''', s(s(s(x''''''))), *(x'''', z''''))
F(x'''', s(s(s(s(x'0''')))), z'''') -> F(x'''', s(s(s(x'0'''))), *(x'''', z''''))
F(s(x'''''), s(s(s(y''''))), z'''') -> F(s(x'''''), s(s(y'''')), *(s(x'''''), z''''))
F(s(s(y''''''')), s(s(s(y''''))), z'''') -> F(s(s(y''''''')), s(s(y'''')), *(s(s(y''''''')), z''''))
F(s(y''''''), s(s(s(s(x''''')))), z'''') -> F(s(y''''''), s(s(s(x'''''))), *(s(y''''''), z''''))
F(x'''', s(s(s(s(x''''')))), z''') -> F(x'''', s(s(s(x'''''))), *(x'''', z'''))
F(x''', s(s(s(s(s(x'''''))))), z''') -> F(x''', s(s(s(s(x''''')))), *(x''', z'''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 23

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(x''', s(s(s(s(s(x'''''))))), z''') -> F(x''', s(s(s(s(x''''')))), *(x''', z'''))
F(x'''', s(s(s(s(x''''')))), z''') -> F(x'''', s(s(s(x'''''))), *(x'''', z'''))
F(s(y''''''), s(s(s(s(x''''')))), z'''') -> F(s(y''''''), s(s(s(x'''''))), *(s(y''''''), z''''))
F(s(s(y''''''')), s(s(s(y''''))), z'''') -> F(s(s(y''''''')), s(s(y'''')), *(s(s(y''''''')), z''''))
F(s(x'''''), s(s(s(y''''))), z'''') -> F(s(x'''''), s(s(y'''')), *(s(x'''''), z''''))
F(x'''', s(s(s(s(x'0''')))), z'''') -> F(x'''', s(s(s(x'0'''))), *(x'''', z''''))
F(x'''', s(s(s(s(x'''''')))), z'''') -> F(x'''', s(s(s(x''''''))), *(x'''', z''''))
F(s(y''''), s(s(s(x''''))), z''') -> F(s(y''''), s(s(x'''')), *(s(y''''), z'''))
F(s(s(y''''')), s(s(y''')), z''') -> F(s(s(y''''')), s(y'''), *(s(s(y''''')), z'''))
F(s(x'''), s(s(y''')), z''') -> F(s(x'''), s(y'''), *(s(x'''), z'''))
F(x'''', s(s(s(y''''))), z'''') -> F(x'''', s(s(y'''')), *(x'''', z''''))
F(s(y'), s(s(x'')), z) -> F(+(*(s(y'), y'), s(y')), s(half(x'')), z)
F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(x'''', s(s(s(x'''''))), z''') -> F(x'''', s(s(x''''')), *(x'''', z'''))
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)
F(x, s(s(s(s(x''')))), z) -> F(*(x, x), s(s(half(x'''))), z)
F(x, s(s(s(0))), z) -> F(*(x, x), s(0), z)
F(x, s(s(0)), z) -> F(*(x, x), s(0), z)
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(0, s(s(x'')), z) -> F(0, s(half(x'')), z)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(x, s(s(0)), z) -> F(*(x, x), s(0), z)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 24

                 ↳Forward Instantiation Transformation

Dependency Pairs:

F(x'''', s(s(s(s(x''''')))), z''') -> F(x'''', s(s(s(x'''''))), *(x'''', z'''))
F(s(y''''''), s(s(s(s(x''''')))), z'''') -> F(s(y''''''), s(s(s(x'''''))), *(s(y''''''), z''''))
F(s(s(y''''''')), s(s(s(y''''))), z'''') -> F(s(s(y''''''')), s(s(y'''')), *(s(s(y''''''')), z''''))
F(s(x'''''), s(s(s(y''''))), z'''') -> F(s(x'''''), s(s(y'''')), *(s(x'''''), z''''))
F(x'''', s(s(s(s(x'0''')))), z'''') -> F(x'''', s(s(s(x'0'''))), *(x'''', z''''))
F(s(y''''), s(s(s(x''''))), z''') -> F(s(y''''), s(s(x'''')), *(s(y''''), z'''))
F(s(s(y''''')), s(s(y''')), z''') -> F(s(s(y''''')), s(y'''), *(s(s(y''''')), z'''))
F(s(x'''), s(s(y''')), z''') -> F(s(x'''), s(y'''), *(s(x'''), z'''))
F(x'''', s(s(s(s(x'''''')))), z'''') -> F(x'''', s(s(s(x''''''))), *(x'''', z''''))
F(0, s(s(x'')), z) -> F(0, s(half(x'')), z)
F(x'''', s(s(s(y''''))), z'''') -> F(x'''', s(s(y'''')), *(x'''', z''''))
F(s(y'), s(s(x'')), z) -> F(+(*(s(y'), y'), s(y')), s(half(x'')), z)
F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(x'''', s(s(s(x'''''))), z''') -> F(x'''', s(s(x''''')), *(x'''', z'''))
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)
F(x, s(s(s(s(x''')))), z) -> F(*(x, x), s(s(half(x'''))), z)
F(x, s(s(s(0))), z) -> F(*(x, x), s(0), z)
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(x''', s(s(s(s(s(x'''''))))), z''') -> F(x''', s(s(s(s(x''''')))), *(x''', z'''))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(x, s(s(s(0))), z) -> F(*(x, x), s(0), z)

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 25

                 ↳Polynomial Ordering

Dependency Pairs:

F(x''', s(s(s(s(s(x'''''))))), z''') -> F(x''', s(s(s(s(x''''')))), *(x''', z'''))
F(s(y''''''), s(s(s(s(x''''')))), z'''') -> F(s(y''''''), s(s(s(x'''''))), *(s(y''''''), z''''))
F(s(s(y''''''')), s(s(s(y''''))), z'''') -> F(s(s(y''''''')), s(s(y'''')), *(s(s(y''''''')), z''''))
F(s(x'''''), s(s(s(y''''))), z'''') -> F(s(x'''''), s(s(y'''')), *(s(x'''''), z''''))
F(x'''', s(s(s(s(x'0''')))), z'''') -> F(x'''', s(s(s(x'0'''))), *(x'''', z''''))
F(s(y''''), s(s(s(x''''))), z''') -> F(s(y''''), s(s(x'''')), *(s(y''''), z'''))
F(s(s(y''''')), s(s(y''')), z''') -> F(s(s(y''''')), s(y'''), *(s(s(y''''')), z'''))
F(s(x'''), s(s(y''')), z''') -> F(s(x'''), s(y'''), *(s(x'''), z'''))
F(x'''', s(s(s(s(x'''''')))), z'''') -> F(x'''', s(s(s(x''''''))), *(x'''', z''''))
F(0, s(s(x'')), z) -> F(0, s(half(x'')), z)
F(x'''', s(s(s(y''''))), z'''') -> F(x'''', s(s(y'''')), *(x'''', z''''))
F(s(y'), s(s(x'')), z) -> F(+(*(s(y'), y'), s(y')), s(half(x'')), z)
F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(x'''', s(s(s(x'''''))), z''') -> F(x'''', s(s(x''''')), *(x'''', z'''))
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)
F(x, s(s(s(s(x''')))), z) -> F(*(x, x), s(s(half(x'''))), z)
F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(x'''', s(s(s(s(x''''')))), z''') -> F(x'''', s(s(s(x'''''))), *(x'''', z'''))

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

F(x''', s(s(s(s(s(x'''''))))), z''') -> F(x''', s(s(s(s(x''''')))), *(x''', z'''))
F(s(y''''''), s(s(s(s(x''''')))), z'''') -> F(s(y''''''), s(s(s(x'''''))), *(s(y''''''), z''''))
F(s(s(y''''''')), s(s(s(y''''))), z'''') -> F(s(s(y''''''')), s(s(y'''')), *(s(s(y''''''')), z''''))
F(s(x'''''), s(s(s(y''''))), z'''') -> F(s(x'''''), s(s(y'''')), *(s(x'''''), z''''))
F(x'''', s(s(s(s(x'0''')))), z'''') -> F(x'''', s(s(s(x'0'''))), *(x'''', z''''))
F(s(y''''), s(s(s(x''''))), z''') -> F(s(y''''), s(s(x'''')), *(s(y''''), z'''))
F(s(s(y''''')), s(s(y''')), z''') -> F(s(s(y''''')), s(y'''), *(s(s(y''''')), z'''))
F(s(x'''), s(s(y''')), z''') -> F(s(x'''), s(y'''), *(s(x'''), z'''))
F(x'''', s(s(s(s(x'''''')))), z'''') -> F(x'''', s(s(s(x''''''))), *(x'''', z''''))
F(0, s(s(x'')), z) -> F(0, s(half(x'')), z)
F(x'''', s(s(s(y''''))), z'''') -> F(x'''', s(s(y'''')), *(x'''', z''''))
F(s(y'), s(s(x'')), z) -> F(+(*(s(y'), y'), s(y')), s(half(x'')), z)
F(s(y''''), s(s(s(x'''))), z'') -> F(s(y''''), s(s(x''')), *(s(y''''), z''))
F(s(s(y''''')), s(s(y'')), z'') -> F(s(s(y''''')), s(y''), *(s(s(y''''')), z''))
F(s(x'''), s(s(y'')), z'') -> F(s(x'''), s(y''), *(s(x'''), z''))
F(x'''', s(s(s(x'''''))), z''') -> F(x'''', s(s(x''''')), *(x'''', z'''))
F(s(y''), s(s(x')), z) -> F(+(*(s(y''), y''), s(y'')), s(half(x')), z)
F(x, s(s(s(s(x''')))), z) -> F(*(x, x), s(s(half(x'''))), z)
F(x', s(s(s(x'0'))), z'') -> F(x', s(s(x'0')), *(x', z''))
F(x'', s(s(s(x''''))), z'') -> F(x'', s(s(x'''')), *(x'', z''))
F(0, s(s(x')), z) -> F(0, s(half(x')), z)
F(x'''', s(s(s(s(x''''')))), z''') -> F(x'''', s(s(s(x'''''))), *(x'''', z'''))

Additionally, the following usable rules for innermost can be oriented:

*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(*(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

POL(half(x₁)) = x₁

POL(+(x₁, x₂)) = 0

POL(F(x₁, x₂, x₃)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

       →DP Problem 5

         ↳Nar

           →DP Problem 18

             ↳Nar

...

               →DP Problem 26

                 ↳Dependency Graph

Dependency Pairs:

F(s(s(y''')), s(y), z) -> F(+(+(*(s(s(y''')), y'''), s(s(y'''))), s(s(y'''))), half(s(y)), z)
F(s(0), s(y), z) -> F(+(0, s(0)), half(s(y)), z)

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
if(true, x, y) -> true
if(false, x, y) -> false
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))
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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:17 minutes

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