Simplifying compound propositions

Question

I am struggling greatly with simplifying compound statements within propositional calculus. I have read textbooks and watched YouTube tutorials, but they always deal with the same (very simple) problems. Whilst these are somewhat helpful in explaining things like DeMorgan's theorem, and expanding brackets, etc. I am struggling when simplifying longer statements. For all I know the statement is as simple as possible already, and that's why my efforts fail...

I have to prove logical validity for the statement below, and I have achieved that using a 64 row truth table, but feel there must be another way (by simplifying the statement).

(A ⇒ F ∧ L) ∧ (F ⇒ M ∧ S) ∧ (M ⇒ R) ⇒ (A ⇒ L ∧ R)

I get this far and then am lost:

¬((¬A ∨ (F ∧ L)) ∧ (¬F ∨ (M ∧ S)) ∧ (¬M ∨ R)) ∨ (¬A ∨ (L ∨ R))

I feel it might be worth mentioning that I have previously gone much further than that step, but seem to always end up with something not equivalent to the original statement, so I'm definitely doing something wrong!

⇒ = IMPLIES
¬ = NOT
∧ = AND
∨ = OR

THE SOLUTION

(Thank you Zar)

Take the first part of the statement (before the conclusion) and multiply out the brackets.

(A ⇒ F ∧ L) ∧ (F ⇒ M ∧ S) ∧ (M ⇒ R) **becomes**
(A ⇒ F ∧ A ⇒ L) ∧ (F ⇒ M ∧ F ⇒ S) ∧ (M ⇒ R)

We then also know that A ⇒ F, F ⇒ M, and M ⇒ R. So, A ⇒ R and we can get rid of F and M from the statement

A ⇒ L stays the same, as there is no way to simplify this.

This means that (So Far, A ⇒ R ∧ A ⇒ L).

A ⇒ F and F ⇒ S, so A ⇒ S

This leaves A ⇒ R and A ⇒ L and A ⇒ S (A ⇒ R ∧ A ⇒ L ∧ A ⇒ S).

Factorise the brackets to get:
A ⇒ (R ∧ L ∧ S)

Finally, add this back at the start of the statement to get our simplified compound statement:
(A ⇒ (R ∧ L ∧ S)) ⇒ (A ⇒ L ∧ R)
You can actually take out the S, as it's not necessary to prove validity, and swap the R ∧ L around to get L ∧ R, and all of a sudden both sides match perfectly and it's easy to see at a glance that this is a tautology.
(A ⇒ L ∧ R) ⇒ (A ⇒ L ∧ R)

Answer 1

From $A\Rightarrow F\wedge L$ you have $A \Rightarrow F$ and $A \Rightarrow L$.

From $F\Rightarrow M\wedge S$ you have $F\Rightarrow M$.

From $F\Rightarrow M$ and $M\Rightarrow R$ you have $F\Rightarrow R$.

From $A\Rightarrow F$ and $F\Rightarrow R$ you have $A\Rightarrow R$.

And now, from $A\Rightarrow R$ and $A \Rightarrow L$, you have $A\Rightarrow L\wedge R$.

Answer 2

Using SymPy Live,

>>> A, F, L, M, R, S = symbols('A F L M R S')
>>> Phi = ((A >> (F & L)) & (F >> (M & S)) & (M >> R)) >> (A >> (L & R))
>>> simplify(Phi)
True

What is the CNF?

>>> to_cnf(Phi)
And(Or(A, F, L, M, Not(A)), Or(A, F, L, Not(A), Not(R)), Or(A, F, M, R, Not(A)), Or(A, F, R, Not(A), Not(R)), Or(A, L, M, Not(A), Not(M), Not(S)), Or(A, L, Not(A), Not(M), Not(R), Not(S)), Or(A, M, R, Not(A), Not(M), Not(S)), Or(A, R, Not(A), Not(M), Not(R), Not(S)), Or(F, L, M, Not(A), Not(F), Not(L)), Or(F, L, Not(A), Not(F), Not(L), Not(R)), Or(F, M, R, Not(A), Not(F), Not(L)), Or(F, R, Not(A), Not(F), Not(L), Not(R)), Or(L, M, Not(A), Not(F), Not(L), Not(M), Not(S)), Or(L, Not(A), Not(F), Not(L), Not(M), Not(R), Not(S)), Or(M, R, Not(A), Not(F), Not(L), Not(M), Not(S)), Or(R, Not(A), Not(F), Not(L), Not(M), Not(R), Not(S)))

What is the DNF?

>>> to_dnf(Phi)
Or(And(A, Not(F)), And(A, Not(L)), And(F, Not(M)), And(F, Not(S)), And(L, R), And(M, Not(R)), Not(A))