f(g(

p(0) -> g(0)

g(s(p(

R

↳Dependency Pair Analysis

F(g(x), g(y)) -> F(p(f(g(x), s(y))), g(s(p(x))))

F(g(x), g(y)) -> P(f(g(x), s(y)))

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

F(g(x), g(y)) -> G(s(p(x)))

F(g(x), g(y)) -> P(x)

P(0) -> G(0)

Furthermore,

R

↳DPs

→DP Problem 1

↳Narrowing Transformation

**F(g( x), g(y)) -> F(p(f(g(x), s(y))), g(s(p(x))))**

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

p(0) -> g(0)

g(s(p(x))) -> p(x)

innermost

On this DP problem, a Narrowing SCC transformation can be performed.

As a result of transforming the rule

two new Dependency Pairs are created:

F(g(x), g(y)) -> F(p(f(g(x), s(y))), g(s(p(x))))

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

F(g(0), g(y)) -> F(p(f(g(0), s(y))), g(s(g(0))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Narrowing Transformation

**F(g(0), g( y)) -> F(p(f(g(0), s(y))), g(s(g(0))))**

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

p(0) -> g(0)

g(s(p(x))) -> p(x)

innermost

On this DP problem, a Narrowing SCC transformation can be performed.

As a result of transforming the rule

no new Dependency Pairs are created.

F(g(0), g(y)) -> F(p(f(g(0), s(y))), g(s(g(0))))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Nar

→DP Problem 2

↳Nar

...

→DP Problem 3

↳Remaining Obligation(s)

The following remains to be proven:

**F(g( x''), g(y)) -> F(p(f(g(x''), s(y))), p(x''))**

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

p(0) -> g(0)

g(s(p(x))) -> p(x)

innermost

Duration:

0:00 minutes