exp(

exp(

*(0,

*(s(

-(0,

-(

-(s(

R

↳Dependency Pair Analysis

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

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

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(exp(x)_{1}, x_{2})= 1 _{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(*'(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

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

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

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(exp(x)_{1}, x_{2})= 1 _{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(-'(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

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

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

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(exp(x)_{1}, x_{2})= 1 _{ }^{ }_{ }^{ }POL(EXP(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(*(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(+(x)_{1}, x_{2})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

exp(x, 0) -> s(0)

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

*(0,y) -> 0

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

-(0,y) -> 0

-(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes