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.

`   R`
`     ↳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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`

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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(-'(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 6`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`

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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(*'(x1, x2)) =  x2 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`

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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`           →DP Problem 8`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polynomial Ordering`
`       →DP Problem 5`
`         ↳Polo`

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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS 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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`           →DP Problem 9`
`             ↳Dependency Graph`
`       →DP Problem 5`
`         ↳Polo`

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polynomial Ordering`

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

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 10`
`             ↳Narrowing Transformation`

Dependency Pair:

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Polynomial Ordering`

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)

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:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Dependency Graph`

Dependency Pairs:

F(s(y''), s(y), z) -> F(+(*(s(y''), y''), s(y'')), half(s(y)), z)
F(0, s(y), z) -> F(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 the DP problem was split into 1 DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`       →DP Problem 3`
`         ↳Polo`
`       →DP Problem 4`
`         ↳Polo`
`       →DP Problem 5`
`         ↳Polo`
`           →DP Problem 10`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pair:

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

Innermost Termination of R could not be shown.
Duration:
0:01 minutes