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

**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)

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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 2

↳Dependency Graph

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)

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes