Question 188838: I saw that people can solve logic proofs on this site. I've looked everywhere for help. Could I get help with these two?
1. p v (q & r)
2. ~r
3. p -> (s -> ~t) /:. (therefore) ~(s & t)
AND
1. (p <--> q) -> s
2. ~(~r -> t)
3. ~q v ~s /:. (t v p) -> (~t & ~q)
I would write what I have so far but I know that it's really wrong and it just confused me.
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website! 1. p v (q & r)
2. ~r
3. p -> (s -> ~t)
:. (therefore) ~(s & t)
Write the conjunction of premises 1 and 2:
[p v (q & r)] & ~r
Distribute inside the bracket:
[(p v q) & (p v r)] & ~r
Use the associative law to move the bracket:
(p v q) & [(p v r) & ~r]
Use the associative law inside the bracket
to move the parentheses:
(p v q) & [(p v (r & ~r)]
(r & ~r) is a contradiction so we replace it by F
(p v q) & [p v F]
Us the distributive law in revers to "factor" out " p v "
p v (q v F)
F is the identity for v so we can replace p by p v F
(p v q) & (p v F)
Use the distributive law to factor out " p v "
p v (q & F)
Since F is the annihilator for & we can replace (q & F) by F
p v F
Since F is the identity for v we can replace that by p
p
Now we take the conjunction of this with premise 3.
p & [p -> (s -> ~t)]
Since x -> y is equivalent to ~x v y, we
can write the bracket thusly:
p & [~p v (s -> ~t)]
Use the distributive law:
(p & ~p) v [p & (s -> ~t)]
(p & ~p) is a contradiction so we can replace it by F
F v [p & (s -> ~t)]
Since F is the identity for v, we can eliminate the F v
p & (s -> ~t)
Use the fact that x & y -> y, so we can eliminate " p & "
s -> ~t
Since x -> y is equivalent to ~x v y, we
can write that as
~s v ~t
We can use DeMorgan's law to "factor" out ~ and get
~(s & t)
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