f(

f(s(

f(c(

g(

g(

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↳Dependency Pair Analysis

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

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

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

Furthermore,

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↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Remaining

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

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

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

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

g(x,y) ->x

g(x,y) ->y

The following dependency pair can be strictly oriented:

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

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})= 0 _{ }^{ }_{ }^{ }POL(s(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2}, x_{3})= x _{1}_{ }^{ }

resulting in one new DP problem.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

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

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

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

g(x,y) ->x

g(x,y) ->y

Using the Dependency Graph resulted in no new DP problems.

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↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Remaining Obligation(s)

The following remains to be proven:

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

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

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

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

g(x,y) ->x

g(x,y) ->y

Duration:

0:00 minutes