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

↳Forward Instantiation Transformation

**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

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

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

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

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

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

On this DP problem, a Forward Instantiation SCC transformation can be performed.

As a result of transforming the rule

one new Dependency Pair is created:

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

H(g(g(g(x'''',y''''),y''0),y)) -> H(g(g(x'''',y''''),y''0))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 3

↳Polynomial Ordering

**H(g(g(g( x'''', y''''), y''0), y)) -> H(g(g(x'''', y''''), y''0))**

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(g(g(x'''',y''''),y''0),y)) -> H(g(g(x'''',y''''),y''0))

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

g(0, 1) -> 0

g(0, 1) -> 1

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1}, x_{2})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(H(x)_{1})= 1 + x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳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