A community for students.
Here's the question you clicked on:
 0 viewing
satellite73
 4 years ago
Fewest number of steps, no truth tables
show
\[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] is a tautology
satellite73
 4 years ago
Fewest number of steps, no truth tables show \[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] is a tautology

This Question is Closed

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0fewest steps would be the wolf I think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i know it is true and i can see why, but i cannot make the steps

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0p>r = pvr q>r = qvr

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't want a truth table

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0these are identities and operations

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wrote it out twice and got lost in the damned ps and qs had \[[(p\lor q)\land(\lnot p\lor r)\land (\lnot q \lor r)]\rightarrow r\] then \[\lnot[(p\lor q)\land(\lnot p\lor r)\land (\lnot q \lor r)]\lor r\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0(pvq)n(pvr)n(qvr) v r (pnq) v (pnr) v (qnr) v r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[[\lnot(p\lor q)\lor \lnot(\lnot p\lor r)\lor \lnot(\lnot q \lor r)]\lor r\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0(pnq) v (pnr) v (qnr) v r (pnq) v (pnr) v (qvr) n (rvr) =r (pnq) v (pnr) v (qvr) n r (pnq) v (r) v (qvr) (pnq) v (rvq) v (rvr) =r (pnq) v (rvq) v r (pnq) v (rvq) so far

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok just about where you are \[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0(pnq) v r v q (pv q) n (qv q) v r =q (p v q v r ) n q = q perhaps

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0thats not a T tho is it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am good to here \[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\] and i can move these around any way i like, bu i cannot seem to get rid of anything

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh maybe i can factor a not r out of the last two terms before the "or r"??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe \[(p\lor q)\land \lnot r\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0since you got all vs inbetween the [ ] seem rather useless

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so \[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\] \[[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0distribute your vr threw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0looks like i should be close right??

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0perhaps (pnq) v [(pvqvr) n (rvr)] =r (pnq) v [(pvqvr) n r] (pnq) v (pnr) v(qnr) v (rnr) maybe?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0that might be a cirle lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aaaaaaaaaaaaaaaaaaaaaargh

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)\lor r\] \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land (\lnot r\lor r)\] \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land r\] ugh is right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i think i may have it please check

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0starting here \[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\lnot p\land \lnot q)\lor r \lor (( p\lor q)\land \lnot r)\] \[(\lnot p\land \lnot q)\lor (r\lor (p\lor q) \land (r \lor \lnot r)\]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\lnot p\land \lnot q)\lor (r\lor (p\lor q)) \land T\]\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[((\lnot p \land \lnot q)\lor r \lor (p\lor q)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lnot( p \lor q) \lor r \lor (p \lor q)\] \[\lnot (p\lor q) \lor (p\lor q) \lor r)\] \[T\lor r\] TTTTTTTTTTTTTTTTTTTTTTTTTTT

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would it be easier with truth table?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe? probably took 8 too many steps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i wanted to do it without truth tables and got all messed up

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0you could just do this with words

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok zarkon, lets see 2 steps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i feel this way about all this nonsense lots of symbols for the completely obvious

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as a professor i once had (who sadly recently died) "just because it is obvious, doesn't mean you can't prove it"

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0\[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] if r is true then the whole thing is true so assume r is false. then for the statement to be false the lhs has to be true this tells us p and q must be false (otherwise the two implications would be false) but then p or q is false and the lhs is false...thus the statement is always true

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.0I should probably write ...this tells us p is false and q is false

phi
 4 years ago
Best ResponseYou've already chosen the best response.0[ (p + q) x (p>r) x (q>r) ] > r ¬ [ (p + q) x (¬p + r) x (¬q + r) ] + r a>b becomes ¬a + b ¬(p+q) + (p x ¬r) + (q x ¬r) + r ¬(a + b) becomes ¬a x ¬b and ¬(a x b) becomes ¬a + ¬b ¬(p+q) + r + ((p + q) x ¬r) commute +r, factor out (p+q) (¬(p+q) + r) + ¬(¬(p + q) + r) ¬a x ¬b becomes ¬(a + b)

phi
 4 years ago
Best ResponseYou've already chosen the best response.0no claims to being the shortest.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes it looks nice and short!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no no you should have see what i wrote!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im too tired to unravel it, beacuse you tend to go back, lol
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.