norm(nil) -> 0

norm(g(

f(

f(

rem(nil,

rem(g(

rem(g(

R

↳Dependency Pair Analysis

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

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

REM(g(x,y), s(z)) -> REM(x,z)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

**NORM(g( x, y)) -> NORM(x)**

norm(nil) -> 0

norm(g(x,y)) -> s(norm(x))

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

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

rem(nil,y) -> nil

rem(g(x,y), 0) -> g(x,y)

rem(g(x,y), s(z)) -> rem(x,z)

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 4

↳Dependency Graph

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

norm(nil) -> 0

norm(g(x,y)) -> s(norm(x))

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

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

rem(nil,y) -> nil

rem(g(x,y), 0) -> g(x,y)

rem(g(x,y), s(z)) -> rem(x,z)

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

→DP Problem 3

↳Polo

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

norm(nil) -> 0

norm(g(x,y)) -> s(norm(x))

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

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

rem(nil,y) -> nil

rem(g(x,y), 0) -> g(x,y)

rem(g(x,y), s(z)) -> rem(x,z)

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 5

↳Dependency Graph

→DP Problem 3

↳Polo

norm(nil) -> 0

norm(g(x,y)) -> s(norm(x))

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

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

rem(nil,y) -> nil

rem(g(x,y), 0) -> g(x,y)

rem(g(x,y), s(z)) -> rem(x,z)

Using the Dependency Graph resulted in no new DP problems.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polynomial Ordering

**REM(g( x, y), s(z)) -> REM(x, z)**

norm(nil) -> 0

norm(g(x,y)) -> s(norm(x))

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

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

rem(nil,y) -> nil

rem(g(x,y), 0) -> g(x,y)

rem(g(x,y), s(z)) -> rem(x,z)

The following dependency pair can be strictly oriented:

REM(g(x,y), s(z)) -> REM(x,z)

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

Used ordering: Polynomial ordering with Polynomial interpretation:

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

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

→DP Problem 3

↳Polo

→DP Problem 6

↳Dependency Graph

norm(nil) -> 0

norm(g(x,y)) -> s(norm(x))

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

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

rem(nil,y) -> nil

rem(g(x,y), 0) -> g(x,y)

rem(g(x,y), s(z)) -> rem(x,z)

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes