-(

-(s(

+(0,

+(s(

*(

*(

f(s(

R

↳Dependency Pair Analysis

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

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

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

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 AFS that need to be oriented.

Used ordering: Homeomorphic Embedding Order with EMB

resulting in one new DP problem.

Used Argument Filtering System:

-'(x,_{1}x) -> -'(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Argument Filtering and Ordering

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Homeomorphic Embedding Order with EMB

resulting in one new DP problem.

Used Argument Filtering System:

+'(x,_{1}x) -> +'(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 6

↳Dependency Graph

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳Argument Filtering and Ordering

→DP Problem 4

↳Rw

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

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 AFS that need to be oriented.

Used ordering: Homeomorphic Embedding Order with EMB

resulting in one new DP problem.

Used Argument Filtering System:

*'(x,_{1}x) -> *'(_{2}x,_{1}x)_{2}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 7

↳Dependency Graph

→DP Problem 4

↳Rw

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rewriting Transformation

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

→DP Problem 8

↳Rewriting Transformation

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

→DP Problem 8

↳Rw

...

→DP Problem 9

↳Rewriting Transformation

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

→DP Problem 8

↳Rw

...

→DP Problem 10

↳Rewriting Transformation

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

→DP Problem 8

↳Rw

...

→DP Problem 11

↳Rewriting Transformation

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

→DP Problem 8

↳Rw

...

→DP Problem 12

↳Rewriting Transformation

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(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:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳AFS

→DP Problem 3

↳AFS

→DP Problem 4

↳Rw

→DP Problem 8

↳Rw

...

→DP Problem 13

↳Remaining Obligation(s)

The following remains to be proven:

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

-(x, 0) ->x

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

+(0,y) ->y

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

*(x, 0) -> 0

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

f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

innermost

Duration:

0:00 minutes