Term Rewriting System R:
[x, y]
even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

EVEN(s(s(x))) -> EVEN(x)
HALF(s(s(x))) -> HALF(x)
PLUS(s(x), y) -> PLUS(x, y)
TIMES(s(x), y) -> IFTIMES(even(s(x)), s(x), y)
TIMES(s(x), y) -> EVEN(s(x))
IFTIMES(true, s(x), y) -> PLUS(times(half(s(x)), y), times(half(s(x)), y))
IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
IFTIMES(true, s(x), y) -> HALF(s(x))
IFTIMES(false, s(x), y) -> PLUS(y, times(x, y))
IFTIMES(false, s(x), y) -> TIMES(x, y)

Furthermore, R contains four SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

EVEN(s(s(x))) -> EVEN(x)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

EVEN(s(s(x))) -> EVEN(x)
one new Dependency Pair is created:

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))
one new Dependency Pair is created:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 6`
`                 ↳Polynomial Ordering`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(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(EVEN(x1)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 5`
`             ↳FwdInst`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

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:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

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:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳FwdInst`
`             ...`
`               →DP Problem 9`
`                 ↳Polynomial Ordering`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

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 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`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`           →DP Problem 8`
`             ↳FwdInst`
`             ...`
`               →DP Problem 10`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

PLUS(s(x), y) -> PLUS(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳FwdInst`
`             ...`
`               →DP Problem 12`
`                 ↳Polynomial Ordering`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

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

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(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(PLUS(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`           →DP Problem 11`
`             ↳FwdInst`
`             ...`
`               →DP Problem 13`
`                 ↳Dependency Graph`
`       →DP Problem 4`
`         ↳Nar`

Dependency Pair:

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Narrowing Transformation`

Dependency Pairs:

IFTIMES(false, s(x), y) -> TIMES(x, y)
IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
TIMES(s(x), y) -> IFTIMES(even(s(x)), s(x), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

TIMES(s(x), y) -> IFTIMES(even(s(x)), s(x), y)
two new Dependency Pairs are created:

TIMES(s(0), y) -> IFTIMES(false, s(0), y)
TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Narrowing Transformation`

Dependency Pairs:

IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
one new Dependency Pair is created:

IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Narrowing Transformation`

Dependency Pairs:

IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)
three new Dependency Pairs are created:

TIMES(s(s(0)), y) -> IFTIMES(true, s(s(0)), y)
TIMES(s(s(s(0))), y) -> IFTIMES(false, s(s(s(0))), y)
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳Narrowing Transformation`

Dependency Pairs:

TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)
TIMES(s(s(s(0))), y) -> IFTIMES(false, s(s(s(0))), y)
TIMES(s(s(0)), y) -> IFTIMES(true, s(s(0)), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)
two new Dependency Pairs are created:

IFTIMES(true, s(s(0)), y) -> TIMES(s(0), y)
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 17`
`                 ↳Instantiation Transformation`

Dependency Pairs:

IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(0))), y) -> IFTIMES(false, s(s(s(0))), y)
IFTIMES(true, s(s(0)), y) -> TIMES(s(0), y)
TIMES(s(s(0)), y) -> IFTIMES(true, s(s(0)), y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IFTIMES(false, s(x), y) -> TIMES(x, y)
three new Dependency Pairs are created:

IFTIMES(false, s(0), y'') -> TIMES(0, y'')
IFTIMES(false, s(s(s(0))), y'') -> TIMES(s(s(0)), y'')
IFTIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 18`
`                 ↳Forward Instantiation Transformation`

Dependency Pairs:

IFTIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

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

IFTIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')
one new Dependency Pair is created:

IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 19`
`                 ↳Polynomial Ordering`

Dependency Pairs:

IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)

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

half(0) -> 0
half(s(s(x))) -> s(half(x))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(IF_TIMES(x1, x2, x3)) =  x2 POL(TIMES(x1, x2)) =  x1 POL(0) =  0 POL(false) =  0 POL(even(x1)) =  0 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(half(x1)) =  x1

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳FwdInst`
`       →DP Problem 3`
`         ↳FwdInst`
`       →DP Problem 4`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 20`
`                 ↳Dependency Graph`

Dependency Pair:

TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)

Rules:

even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes