f(0, 1, g(

g(0, 1) -> 0

g(0, 1) -> 1

h(g(

R

↳Dependency Pair Analysis

F(0, 1, g(x,y),z) -> F(g(x,y), g(x,y), g(x,y), h(x))

F(0, 1, g(x,y),z) -> H(x)

H(g(x,y)) -> H(x)

Furthermore,

R

↳DPs

→DP Problem 1

↳Argument Filtering and Ordering

→DP Problem 2

↳Remaining

**H(g( x, y)) -> H(x)**

f(0, 1, g(x,y),z) -> f(g(x,y), g(x,y), g(x,y), h(x))

g(0, 1) -> 0

g(0, 1) -> 1

h(g(x,y)) -> h(x)

The following dependency pair can be strictly oriented:

H(g(x,y)) -> H(x)

There are no usable rules w.r.t. to the AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(g(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(H(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

Used Argument Filtering System:

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

g(x,_{1}x) -> g(_{2}x,_{1}x)_{2}

R

↳DPs

→DP Problem 1

↳AFS

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Remaining

f(0, 1, g(x,y),z) -> f(g(x,y), g(x,y), g(x,y), h(x))

g(0, 1) -> 0

g(0, 1) -> 1

h(g(x,y)) -> h(x)

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:

**F(0, 1, g( x, y), z) -> F(g(x, y), g(x, y), g(x, y), h(x))**

f(0, 1, g(x,y),z) -> f(g(x,y), g(x,y), g(x,y), h(x))

g(0, 1) -> 0

g(0, 1) -> 1

h(g(x,y)) -> h(x)

Duration:

0:00 minutes