f(

f(

g(empty,

g(cons(

R

↳Removing Redundant Rules

Removing the following rules from

f(a, empty) -> g(a, empty)

where the Polynomial interpretation:

was used.

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

Not all Rules of

R

↳RRRPolo

→TRS2

↳Removing Redundant Rules

Removing the following rules from

g(empty,d) ->d

where the Polynomial interpretation:

was used.

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

Not all Rules of

R

↳RRRPolo

→TRS2

↳RRRPolo

→TRS3

↳Removing Redundant Rules

Removing the following rules from

f(a, cons(x,k)) -> f(cons(x,a),k)

where the Polynomial interpretation:

was used.

_{ }^{ }POL(g(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(cons(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(f(x)_{1}, x_{2})= x _{1}+ 2·x_{2}_{ }^{ }

Not all Rules of

R

↳RRRPolo

→TRS2

↳RRRPolo

→TRS3

↳RRRPolo

...

→TRS4

↳Removing Redundant Rules

Removing the following rules from

g(cons(x,k),d) -> g(k, cons(x,d))

where the Polynomial interpretation:

was used.

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

All Rules of

R

↳RRRPolo

→TRS2

↳RRRPolo

→TRS3

↳RRRPolo

...

→TRS5

↳Dependency Pair Analysis

Duration:

0:00 minutes