a(lambda(

a(lambda(

a(a(

a(

a(

lambda(

R

↳Dependency Pair Analysis

A(lambda(x),y) -> LAMBDA(a(x, 1))

A(lambda(x),y) -> A(x, 1)

A(lambda(x),y) -> LAMBDA(a(x, a(y, t)))

A(lambda(x),y) -> A(x, a(y, t))

A(lambda(x),y) -> A(y, t)

A(a(x,y),z) -> A(x, a(y,z))

A(a(x,y),z) -> A(y,z)

Furthermore,

R

↳DPs

→DP Problem 1

↳Polynomial Ordering

**A(a( x, y), z) -> A(y, z)**

a(lambda(x),y) -> lambda(a(x, 1))

a(lambda(x),y) -> lambda(a(x, a(y, t)))

a(a(x,y),z) -> a(x, a(y,z))

a(x,y) ->x

a(x,y) ->y

lambda(x) ->x

The following dependency pairs can be strictly oriented:

A(lambda(x),y) -> A(y, t)

A(lambda(x),y) -> A(x, a(y, t))

A(lambda(x),y) -> A(x, 1)

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

a(lambda(x),y) -> lambda(a(x, 1))

a(lambda(x),y) -> lambda(a(x, a(y, t)))

a(a(x,y),z) -> a(x, a(y,z))

a(x,y) ->x

a(x,y) ->y

lambda(x) ->x

Used ordering: Polynomial ordering with Polynomial interpretation:

_{ }^{ }POL(t)= 0 _{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(lambda(x)_{1})= 1 + x _{1}_{ }^{ }_{ }^{ }POL(a(x)_{1}, x_{2})= x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(A(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polynomial Ordering

**A(a( x, y), z) -> A(y, z)**

a(lambda(x),y) -> lambda(a(x, 1))

a(lambda(x),y) -> lambda(a(x, a(y, t)))

a(a(x,y),z) -> a(x, a(y,z))

a(x,y) ->x

a(x,y) ->y

lambda(x) ->x

The following dependency pairs can be strictly oriented:

A(a(x,y),z) -> A(y,z)

A(a(x,y),z) -> A(x, a(y,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(t)= 0 _{ }^{ }_{ }^{ }POL(1)= 0 _{ }^{ }_{ }^{ }POL(lambda(x)_{1})= 0 _{ }^{ }_{ }^{ }POL(a(x)_{1}, x_{2})= 1 + x _{1}+ x_{2}_{ }^{ }_{ }^{ }POL(A(x)_{1}, x_{2})= x _{1}_{ }^{ }

resulting in one new DP problem.

R

↳DPs

→DP Problem 1

↳Polo

→DP Problem 2

↳Polo

...

→DP Problem 3

↳Dependency Graph

a(lambda(x),y) -> lambda(a(x, 1))

a(lambda(x),y) -> lambda(a(x, a(y, t)))

a(a(x,y),z) -> a(x, a(y,z))

a(x,y) ->x

a(x,y) ->y

lambda(x) ->x

Using the Dependency Graph resulted in no new DP problems.

Duration:

0:00 minutes