Your next step would be to use the Law of Contrapositives to change p -> q and r -> s into ~q -> ~p and ~s -> ~r.
Since you know ~q ^ ~s already, you're a short step away from your goal.
Nicely done so far!
Anyone good with proposition logics? that can help out?
im stuck on one question and can't figure out the next step.
The question states : Use the rules of inference to prove the following.
(p -> q) ^ (r -> s) ^ [t -> ~(q V s) ^ t => (~p ^ ~r)
note : the " ~ " sign means NOT,
the " ^ " sign means AND,
the " -> " sign means implies/conditional
the " V " sign means OR
the " => " sign means resultant or conclusion is.
I've set up my working up to look like this.
P -> q H1(Hypothesis 1)
r -> s H2(Hypothesis 2)
t -> ~(q V s) H3(Hypothesis 3)
t H4(Hypothesis 4)
~p ^ ~r C (Conclusion)
Step one. I got H4 with H3 and AND both of them >> H4 ^ H3 which gives:
t ^ [t -> ~(q V s)] and since t ^ t cancels out with the modus ponens rules it leaves us with ~(q V s).
So now that H4 and H3 is gone we have H1 and H2 left. This is where i got stuck.
I decided to use de morgan's law to simplify which gave me >> ~(q V s) => ~q ^ ~s and we can call this theory one (1).
So far i've only got up to this part and now i'm stuck could you help me figured out what i need to do next?
Working : t ^ [t -> ~(q V s)] <=> ~(q V s) H4 ^ H3
<=> ~q ^ ~s (1).
Next step = ? please help
Damn... i suck bad...
is (~p -> ~q) ^ (~q ^ ~s) possible? using the Hypothecial Syllogism concept. i know it works for if both of them are conditional but just wondering if this method works?
if it does would it look something like ~p->~s or ~p^~s ._.
if ~p->~s works out then... it would be alot easier because then i would just use Hypothetical syllgism law again for ~p->~s ^ ~s->~r which gives me ~p->~r but... then again it doesnt work out... sorry im really bad with this topic. a little more help please?
Ok i think i got it but could someone check the answer for me? if i'm wrong let me know where i got wrong.
t ^ [t -> ~(q V s)] <=> ~(q V s) ... H4 ^ H3 Modus ponens
<=> ~q ^ ~s (1) ... de morgan's law
p -> q => ~q -> ~p (2) ... Contrapositive
r -> s => ~s -> ~r (3) ... Contrapositive
(~q -> ~p) ^ (~q ^ ~s) .... (2) ^ (1)
=> ~p ^ ~s ... Disjunuctive Syllogism
=> ~s ^ ~p (4) ... Commutativity
(~s -> ~r) ^ (~s ^ ~p) (3) ^ (4)
=> ~r ^ ~p ... Disjunctive Syllogism
=> ~p ^ ~r ... Commutativity <--- Conclusion