f(

f(s(

g(0,

R

↳Dependency Pair Analysis

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

G(0,x) -> G(f(x,x),x)

G(0,x) -> F(x,x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

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

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

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

g(0,x) -> g(f(x,x),x)

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

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

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

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

g(0,x) -> g(f(x,x),x)

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Nar

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

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

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

g(0,x) -> g(f(x,x),x)

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost 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(F(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Nar

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

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

g(0,x) -> g(f(x,x),x)

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Narrowing Transformation

**G(0, x) -> G(f(x, x), x)**

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

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

g(0,x) -> g(f(x,x),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:

G(0,x) -> G(f(x,x),x)

G(0, 0) -> G(s(0), 0)

G(0, s(x'')) -> G(s(f(x'',x'')), s(x''))

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes