f(0) -> true

f(1) -> false

f(s(

if(true,

if(false,

g(s(

g(

R

↳Dependency Pair Analysis

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

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

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

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

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

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

f(0) -> true

f(1) -> false

f(s(x)) -> f(x)

if(true,x,y) ->x

if(false,x,y) ->y

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

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

The following dependency pair can be strictly oriented:

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

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(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

f(0) -> true

f(1) -> false

f(s(x)) -> f(x)

if(true,x,y) ->x

if(false,x,y) ->y

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

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

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

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

f(0) -> true

f(1) -> false

f(s(x)) -> f(x)

if(true,x,y) ->x

if(false,x,y) ->y

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

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

The following dependency pairs can be strictly oriented:

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

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

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

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

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

if(true,x,y) ->x

if(false,x,y) ->y

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(if(x)_{1}, x_{2}, x_{3})= x _{2}+ x_{3}_{ }^{ }_{ }^{ }POL(c(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(false)= 0 _{ }^{ }_{ }^{ }POL(G(x)_{1}, x_{2})= x _{2}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(true)= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(f(x)_{1})= 0 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

f(0) -> true

f(1) -> false

f(s(x)) -> f(x)

if(true,x,y) ->x

if(false,x,y) ->y

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

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

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes