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

↳Argument Filtering and Ordering

→DP Problem 2

↳Remaining

**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))

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Homeomorphic Embedding Order with EMB

resulting in one new DP problem.

Used Argument Filtering System:

F(x) -> F(_{1}x)_{1}

s(x) -> s(_{1}x)_{1}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

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))

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( 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))

innermost

Duration:

0:00 minutes