>>10888841You should be able to get there from that point.
In my previous post, from:
~(p <-> q)
We've derived:
~(p -> q) ? ~(q -> p)
Now rememjber that (p -> q) is equivalent to (~p ? q). So, using de Morgan's law again:
(p & ~q) ? (q & ~p)
Now distribute the ?:
(p ? (q & ~p)) & (~q ? (q & ~p))
Distribute the ? again inside each of the conjuncts of the major conjuction:
[(p ? q) & (p ? ~p)] & [(~q ? q) & (~q ?~p)]
Because of the associativity of conjunction, you don't need all the brackets:
(p ? q) & (p ? ~p) & (~q ? q) & (~q ?~p)
If a conjunction is true, all of its conjuncts must be true; therefore, in particular, we have that these are true:
(p ? q) & (~q ?~p)
Since (p -> q) is equivalent to (~p ? q), it follows that (~p -> q) is equivalent to (p ? q) because of the law of double negation. Similarly, (q -> ~p) is equivalent to (~q ?~p). So you get:
(~p -> q) & (q -> ~p)
Which was to be proved.
