f(0) -> 1

f(s(

g(0,

g(s(

g(s(

+(

+(

R

↳Dependency Pair Analysis

F(s(x)) -> G(x, s(x))

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

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

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

The following dependency pairs can be strictly oriented:

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

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(G(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + 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 4

↳Dependency Graph

f(0) -> 1

f(s(x)) -> g(x, s(x))

g(0,y) ->y

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

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

+(x, 0) ->x

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes