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

↳Polynomial Ordering

→DP Problem 2

↳Polo

**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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(remove(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(REMOVE(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(if(x)_{1}, x_{2}, x_{3})= 0 _{ }^{ }_{ }^{ }POL(=(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(nil)= 0 _{ }^{ }_{ }^{ }POL(.(x)_{1}, x_{2})= 1 + x _{2}_{ }^{ }_{ }^{ }POL(purge(x)_{1})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 3

↳Dependency Graph

→DP Problem 2

↳Polo

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)))

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**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)))

The following dependency pair can be strictly oriented:

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

Additionally, the following rules can be oriented:

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)))

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(remove(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(if(x)_{1}, x_{2}, x_{3})= 0 _{ }^{ }_{ }^{ }POL(PURGE(x)_{1})= x _{1}_{ }^{ }_{ }^{ }POL(=(x)_{1}, x_{2})= 0 _{ }^{ }_{ }^{ }POL(nil)= 0 _{ }^{ }_{ }^{ }POL(.(x)_{1}, x_{2})= 1 _{ }^{ }_{ }^{ }POL(purge(x)_{1})= 1 _{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 4

↳Dependency Graph

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)))

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes