times(

times(

times(

plus(

R

↳Dependency Pair Analysis

TIMES(x, plus(y, 1)) -> PLUS(times(x, plus(y, times(1, 0))),x)

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0)))

TIMES(x, plus(y, 1)) -> PLUS(y, times(1, 0))

TIMES(x, plus(y, 1)) -> TIMES(1, 0)

Furthermore,

R

↳DPs

→DP Problem 1

↳Rewriting Transformation

**TIMES( x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0)))**

times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))),x)

times(x, 1) ->x

times(x, 0) -> 0

plus(x, 0) ->x

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, times(1, 0)))

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, 0))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Rewriting Transformation

**TIMES( x, plus(y, 1)) -> TIMES(x, plus(y, 0))**

times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))),x)

times(x, 1) ->x

times(x, 0) -> 0

plus(x, 0) ->x

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

TIMES(x, plus(y, 1)) -> TIMES(x, plus(y, 0))

TIMES(x, plus(y, 1)) -> TIMES(x,y)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Rw

...

→DP Problem 3

↳Polynomial Ordering

**TIMES( x, plus(y, 1)) -> TIMES(x, y)**

times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))),x)

times(x, 1) ->x

times(x, 0) -> 0

plus(x, 0) ->x

innermost

The following dependency pair can be strictly oriented:

TIMES(x, plus(y, 1)) -> TIMES(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(TIMES(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(plus(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Rw

→DP Problem 2

↳Rw

...

→DP Problem 4

↳Dependency Graph

times(x, plus(y, 1)) -> plus(times(x, plus(y, times(1, 0))),x)

times(x, 1) ->x

times(x, 0) -> 0

plus(x, 0) ->x

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes