f(f(0,

f(g(

R

↳Dependency Pair Analysis

F(f(0,x), 1) -> F(g(f(x,x)),x)

F(f(0,x), 1) -> F(x,x)

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

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

**F(f(0, x), 1) -> F(x, x)**

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

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:

F(f(0,x), 1) -> F(x,x)

F(f(0, g(x'')), 1) -> F(g(x''), g(x''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Forward Instantiation Transformation

**F(f(0, g( x'')), 1) -> F(g(x''), g(x''))**

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

innermost

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

As a result of transforming the rule

three new Dependency Pairs are created:

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

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

F(g(f(0,x'')), 1) -> F(f(0,x''), 1)

F(g(f(0, g(x''''))), 1) -> F(f(0, g(x'''')), 1)

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 3

↳Narrowing Transformation

**F(g(f(0, g( x''''))), 1) -> F(f(0, g(x'''')), 1)**

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

innermost

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

As a result of transforming the rule

one new Dependency Pair is created:

F(f(0,x), 1) -> F(g(f(x,x)),x)

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

**F(f(0, g( x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))**

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

innermost

The following dependency pairs can be strictly oriented:

F(f(0, g(x'')), 1) -> F(g(g(f(x'', g(x'')))), g(x''))

F(f(0, g(x'')), 1) -> F(g(x''), g(x''))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(0)= 0 _{ }^{ }_{ }^{ }POL(g(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(1)= 1 _{ }^{ }_{ }^{ }POL(f(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(F(x)_{1}, x_{2})= x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

**F(g(f(0, x'')), 1) -> F(f(0, x''), 1)**

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 6

↳Polynomial Ordering

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

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

innermost

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳FwdInst

...

→DP Problem 7

↳Dependency Graph

f(f(0,x), 1) -> f(g(f(x,x)),x)

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

innermost

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes