-(

-(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

↳Argument Filtering and Ordering

→DP Problem 2

↳Remaining

**-'(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)))

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 3

↳Dependency Graph

→DP Problem 2

↳Remaining

-(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)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

**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)))

innermost

Duration:

0:00 minutes