g(c(

f(c(s(

f(f(

f(

R

↳Dependency Pair Analysis

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

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

F(f(x)) -> F(d(f(x)))

Furthermore,

R

↳DPs

→DP Problem 1

↳Instantiation Transformation

→DP Problem 2

↳Inst

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

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 3

↳Instantiation Transformation

→DP Problem 2

↳Inst

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

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 3

↳Inst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Inst

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

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 3

↳Inst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Inst

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Instantiation Transformation

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

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Inst

→DP Problem 6

↳Instantiation Transformation

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

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

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

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Inst

→DP Problem 6

↳Inst

...

→DP Problem 7

↳Polynomial Ordering

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

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

The following dependency pair can be strictly oriented:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Inst

→DP Problem 2

↳Inst

→DP Problem 6

↳Inst

...

→DP Problem 8

↳Dependency Graph

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

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

f(f(x)) -> f(d(f(x)))

f(x) ->x

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes