Exercise 2.2b from A New Introduction to Modal Logic

Question

I am kind of stuck on page 48 on exercise 2.2b in Hughes and Cresswell; In case you don't have the book at hand here's the question:

Let K** be K but with N and K replaced by

LT: L( p→ p),

R*: ⊢ a → b ⇒ ⊢ La → Lb

and

K2*: (Lp&Lq) → L(p&q).

Show that K and K** have the same theorems.

What I thought is that in 2.2a we have showed that LT+R* ⊢ N, so it's enough to show that LT+R*+N+K2* ⊢ K.

Now I thought of old trick of math to look from the end back to the start of the proof, i.e:

We need to prove:

L(p → q) → (Lp → Lq)

But I don't see how to use the other rules and axioms, any hints?

Answer 1

I contacted one of the authors of the textbook, Max Cresswell and he kindly suggested me a solution to the problem.

All I need to prove is that K follows from K**.

So here's a proof:

1. ((p->q)&p)->q PC axiom.

2. L((p->q)&p)->Lq 1,R*.

3. (L(p->q)&Lp)->L((p->q)&p) K2*.

4. L(p->q)->(Lp->L((p->q)&p)) 3,PC.

5. L(p->q)->(Lp->Lq) 2,4, PC.

Where PC is axiom of propositional logic, and in 5 I used a sort of hypothetical syllogism.