half(0) -> 0

half(s(0)) -> 0

half(s(s(

s(log(0)) -> s(0)

log(s(

R

↳Dependency Pair Analysis

HALF(s(s(x))) -> S(half(x))

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

S(log(0)) -> S(0)

LOG(s(x)) -> S(log(half(s(x))))

LOG(s(x)) -> LOG(half(s(x)))

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

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

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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 3

↳Polynomial Ordering

→DP Problem 2

↳Nar

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

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

The following dependency pair can be strictly oriented:

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

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

s(log(0)) -> s(0)

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(HALF(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(log(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳Polo

...

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Nar

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Narrowing Transformation

**LOG(s( x)) -> LOG(half(s(x)))**

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

On this DP problem, a Narrowing SCC transformation can be performed.

As a result of transforming the rule

two new Dependency Pairs are created:

LOG(s(x)) -> LOG(half(s(x)))

LOG(s(0)) -> LOG(0)

LOG(s(s(x''))) -> LOG(s(half(x'')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 5

↳Narrowing Transformation

**LOG(s(s( x''))) -> LOG(s(half(x'')))**

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

On this DP problem, a Narrowing SCC transformation can be performed.

As a result of transforming the rule

three new Dependency Pairs are created:

LOG(s(s(x''))) -> LOG(s(half(x'')))

LOG(s(s(0))) -> LOG(s(0))

LOG(s(s(s(0)))) -> LOG(s(0))

LOG(s(s(s(s(x'))))) -> LOG(s(s(half(x'))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 5

↳Nar

...

→DP Problem 6

↳Polynomial Ordering

**LOG(s(s(s(s( x'))))) -> LOG(s(s(half(x'))))**

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

The following dependency pair can be strictly oriented:

LOG(s(s(s(s(x'))))) -> LOG(s(s(half(x'))))

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

s(log(0)) -> s(0)

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(log(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(half(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(LOG(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Nar

→DP Problem 5

↳Nar

...

→DP Problem 7

↳Dependency Graph

half(0) -> 0

half(s(0)) -> 0

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

s(log(0)) -> s(0)

log(s(x)) -> s(log(half(s(x))))

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes