purge(nil) -> nil

purge(.(

remove(

remove(

R

↳Dependency Pair Analysis

PURGE(.(x,y)) -> PURGE(remove(x,y))

PURGE(.(x,y)) -> REMOVE(x,y)

REMOVE(x, .(y,z)) -> REMOVE(x,z)

Furthermore,

R

↳DPs

→DP Problem 1

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

**REMOVE( x, .(y, z)) -> REMOVE(x, z)**

purge(nil) -> nil

purge(.(x,y)) -> .(x, purge(remove(x,y)))

remove(x, nil) -> nil

remove(x, .(y,z)) -> if(=(x,y), remove(x,z), .(y, remove(x,z)))

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:

REMOVE(x, .(y,z)) -> REMOVE(x,z)

REMOVE(x'', .(y, .(y'',z''))) -> REMOVE(x'', .(y'',z''))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳Forward Instantiation Transformation

→DP Problem 2

↳Nar

**REMOVE( x'', .(y, .(y'', z''))) -> REMOVE(x'', .(y'', z''))**

purge(nil) -> nil

purge(.(x,y)) -> .(x, purge(remove(x,y)))

remove(x, nil) -> nil

remove(x, .(y,z)) -> if(=(x,y), remove(x,z), .(y, remove(x,z)))

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:

REMOVE(x'', .(y, .(y'',z''))) -> REMOVE(x'', .(y'',z''))

REMOVE(x'''', .(y, .(y''0, .(y'''',z'''')))) -> REMOVE(x'''', .(y''0, .(y'''',z'''')))

The transformation is resulting in one new DP problem:

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 4

↳Polynomial Ordering

→DP Problem 2

↳Nar

**REMOVE( x'''', .(y, .(y''0, .(y'''', z'''')))) -> REMOVE(x'''', .(y''0, .(y'''', z'''')))**

purge(nil) -> nil

purge(.(x,y)) -> .(x, purge(remove(x,y)))

remove(x, nil) -> nil

remove(x, .(y,z)) -> if(=(x,y), remove(x,z), .(y, remove(x,z)))

innermost

The following dependency pair can be strictly oriented:

REMOVE(x'''', .(y, .(y''0, .(y'''',z'''')))) -> REMOVE(x'''', .(y''0, .(y'''',z'''')))

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(REMOVE(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(.(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 3

↳FwdInst

...

→DP Problem 5

↳Dependency Graph

→DP Problem 2

↳Nar

purge(nil) -> nil

purge(.(x,y)) -> .(x, purge(remove(x,y)))

remove(x, nil) -> nil

remove(x, .(y,z)) -> if(=(x,y), remove(x,z), .(y, remove(x,z)))

innermost

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳FwdInst

→DP Problem 2

↳Narrowing Transformation

**PURGE(.( x, y)) -> PURGE(remove(x, y))**

purge(nil) -> nil

purge(.(x,y)) -> .(x, purge(remove(x,y)))

remove(x, nil) -> nil

remove(x, .(y,z)) -> if(=(x,y), remove(x,z), .(y, remove(x,z)))

innermost

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

As a result of transforming the rule

two new Dependency Pairs are created:

PURGE(.(x,y)) -> PURGE(remove(x,y))

PURGE(.(x'', nil)) -> PURGE(nil)

PURGE(.(x'', .(y'',z'))) -> PURGE(if(=(x'',y''), remove(x'',z'), .(y'', remove(x'',z'))))

The transformation is resulting in no new DP problems.

Duration:

0:00 minutes