-(

-(0, s(

-(s(

f(0) -> 0

f(s(

g(0) -> s(0)

g(s(

R

↳Dependency Pair Analysis

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

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

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

F(s(x)) -> F(x)

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

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

G(s(x)) -> G(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

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

-(x, 0) ->x

-(0, s(y)) -> 0

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

f(0) -> 0

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

g(0) -> s(0)

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

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

-(x, 0) ->x

-(0, s(y)) -> 0

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

f(0) -> 0

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

g(0) -> s(0)

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**G(s( x)) -> G(x)**

-(x, 0) ->x

-(0, s(y)) -> 0

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

f(0) -> 0

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

g(0) -> s(0)

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

The following dependency pairs can be strictly oriented:

G(s(x)) -> G(x)

F(s(x)) -> F(x)

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

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

g(0) -> s(0)

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

f(0) -> 0

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

-(x, 0) ->x

-(0, s(y)) -> 0

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(G(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(-(x)_{1}, x_{2})= x _{1}_{ }^{ }_{ }^{ }POL(f(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

**G(s( x)) -> F(g(x))**

-(x, 0) ->x

-(0, s(y)) -> 0

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

f(0) -> 0

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

g(0) -> s(0)

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes