Is ((f f) (g g)) reduced differently in AOR and NOR?

Question

How is ((f f) (g g)) reduced in both applicative order reduction and normal order reduction? do both reduce the statement in the same way?

Answer 1

Assume "reduce" means "Beta-Reduct".
I'm not going to be rigorous and introduice formals definitions, because of this is not the question. Before, i need to clarify some terms :

• A beta-contraction (writes ~> here) means a single step of the evaluation.
  For example (λx.λy.xy)(a)(b) ~> (λy.ay)(b) forall terms a, b
• A beta-reduction (writes => here) means a complete evaluation.
  For example (λx.λy.xy)(a)(b) => ab forall atoms a, b
• A term is in a beta-normal form if it's no contain redexes.
  For example λx.x is a beta-normal form, but (λx.x)y has a beta-normal form (y) iff y has a beta-normal form. An other important point : A term A has a unique beta-normal form regardless wich strategy we apply to reduce A.
• Two terms are beta-equals (writes == here) if their beta-normal forms are alpha-equivalent For example (λx.λy.xy)(a)(b) == (λy.ay)(b)

In the case of the AOR strategy, to reduce a term A, we beta-contract the rightmost innermost redex until we get a beta-normal form.
More visually :

>> (λx.x)((λy.y)((λz.z)a))
~> (λx.x)((λy.y)a)
~> (λx.x)a
~> a

In the case of the NOR strategy, to reduce a term A, we beta-contract the leftmost outermost redex until we get a beta-normal form More visually :

>> (λx.x)((λy.y)((λz.z)a))
~> (λy.y)((λz.z)a)
~> (λz.z)a
~> a

So, in your case, with f f => A, g g => B
AOR :

>> (f f)(g g)
~> (f f)B
~> AB

NOR :

>> (f f)(g g)
~> A(g g)
~> AB

It's not very rigorous, but the idea is clear. And, of course, (f f)(g g) == (f f)(g g)