Equivalent formulations of the axiom K in modal logic

Question

In modal logic, the axiom K can be equivalently formulated as:

(1) $\Box(p\rightarrow q)\rightarrow (\Box p\rightarrow\Box q)$

(2) $\Box(p\wedge q)\leftrightarrow (\Box p\wedge\Box q)$

(3) $\Diamond(p\vee q)\leftrightarrow (\Diamond p\vee\Diamond q)$

The equivalence between (2) and (3) is just a matter of applying De Morgan's laws. I am more troubled however with the equivalences between (1) and (2), or (1) and (3).

Could anybody please help? Thanks a lot in advance.

Answer 1

For : (1) implies (2) :

1) $(p \land q) \rightarrow p$ --- tautology

2) $\square ( (p \land q) \rightarrow p)$ --- from 1) by Rule of Necessitation

3) $\square (p \land q) \rightarrow \square p$ --- from 2) by Ax K and modus ponens

4) $(p \land q) \rightarrow q$ --- tautology

5) $\square (p \land q) \rightarrow \square q$ --- from 4) as in 2)-3) by Ax K

6) $(p \rightarrow q) \rightarrow ((p \rightarrow r) \rightarrow (p \rightarrow (q \land r)))$ --- tautology

7) $\square (p \land q) \rightarrow (\square p \land \square q)$ --- from 6) with 3) and 5) by modus ponens twice.

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1) $p \rightarrow (q \rightarrow (p \land q))$ --- tautology

2) $\square (p \rightarrow (q \rightarrow (p \land q)))$ --- from 1) by Rule of Necessitation

3) $\square p \rightarrow \square (q \rightarrow (p \land q))$ --- from 2) by Ax K and modus ponens

4) $\square (q \rightarrow (p \land q)) \rightarrow (\square q \rightarrow \square (p \land q))$ --- from Ax K with $q$ in place of $p$ and $p \land q$ in place of $q$

5) $(p \rightarrow q) \rightarrow ((q \rightarrow (r \rightarrow s)) \rightarrow ((p \land r) \rightarrow s))$ --- tautology

6) $(\square p \land \square q) \rightarrow \square (p \land q)$ --- from 5) with 3) and 4) by modus ponens twice.

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Thus :

i) $\square (p \land q) \rightarrow (\square p \land \square q)$

ii) $(\square p \land \square q) \rightarrow \square (p \land q)$

iii) $(p \rightarrow q) \rightarrow ((q \rightarrow p) \rightarrow (p \leftrightarrow q))$ --- tautology

iv) $\square (p \land q) \leftrightarrow (\square p \land \square q)$ --- from iii) with i) and ii) by modus ponens twice.

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In order to prove that (2) implies (1), we have to add a new inference rule :

if $\vdash \alpha \leftrightarrow \beta$, then $\vdash \square \alpha \leftrightarrow \square \beta \ $ (the necessitation of provable equivalents).

1) $(\square p \land \square (p \rightarrow q)) \leftrightarrow \square (p \land (p \rightarrow q))$ --- Ax (2)

2) $(p \land (p \rightarrow q)) \leftrightarrow (p \land q)$ --- tautology

3) $\square(p \land (p \rightarrow q)) \leftrightarrow \square (p \land q)$ --- from 2) by the new rule on equivalents

4) $(p \leftrightarrow q) \rightarrow ((q \leftrightarrow r) \rightarrow (p \leftrightarrow r))$ --- tautology

5) $(\square p \land \square (p \rightarrow q)) \leftrightarrow \square (p \land q)$ --- from 1), 3) and 4) by modus ponens twice

6) $(\square p \land \square (p \rightarrow q)) \leftrightarrow (\square p \land \square q)$ --- from 5), Ax (2) and 4) by modus ponens twice

7) $(\square p \land \square (p \rightarrow q)) \rightarrow (\square p \land \square q)$ --- from 6) and tautological implication

8) $((p \land q) \rightarrow (p \land r)) \rightarrow (q \rightarrow (p \rightarrow r))$ --- tautology

9) $\square (p \rightarrow q) \rightarrow (\square p \rightarrow \square q)$ --- from 7) and 8) by modus ponens.

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You can see G.E. Hughes & M.J. Cresswell, An Introduction to Modal Logic (1968), page 34 and page 124.