Question

Fewest number of steps, no truth tables
show
\[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] is a tautology

Answer 1

fewest steps would be the wolf I think

Answer 2

i know it is true and i can see why, but i cannot make the steps

Answer 3

p>r = -pvr
q>r = -qvr

Answer 4

i don't want a truth table

Answer 5

these are identities and operations

Answer 6

[]>r = -[]vr

Answer 7

wrote 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
\[\lnot[(p\lor q)\land(\lnot p\lor r)\land (\lnot q \lor r)]\lor r\]

Answer 8

-(pvq)-n-(-pvr)-n-(-qvr) v r

(-pn-q) v (pn-r) v (qn-r) v r

Answer 9

\[[\lnot(p\lor q)\lor \lnot(\lnot p\lor r)\lor \lnot(\lnot q \lor r)]\lor r\]

Answer 10

(-pn-q) v (pn-r) v (qn-r) v r

(-pn-q) v (pn-r) v (qvr) n (-rvr)
=r

(-pn-q) v (pn-r) v (qvr)
n r

(-pn-q) v (-r) v (qvr)

(-pn-q) v (-rvq) v (-rvr)
=r

(-pn-q) v (-rvq)
v r

(-pn-q) v (rvq)

so far

Answer 11

ok just about where you are
\[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\]

Answer 12

(-pn-q) v r v q

(-pv q) n (-qv q) v r
=q

(-p v q v r ) n q
= q

perhaps

Answer 13

problem is i want T

Answer 14

thats not a T tho is it

Answer 15

i 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

Answer 16

undistribute maybe

Answer 17

oh maybe i can factor a not r out of the last two terms before the "or r"??

Answer 18

yep

Answer 19

which gets me what?

Answer 20

(pvq) n -r

Answer 21

maybe
\[(p\lor q)\land \lnot r\]

Answer 22

since you got all vs inbetween the [ ] seem rather useless

Answer 23

ok so
\[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\]
\[[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

Answer 24

distribute your vr threw

Answer 25

\[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

Answer 26

looks like i should be close right??

Answer 27

perhaps

(-pnq) v [(pvqvr) n (-rvr)]
=r

(-pnq) v [(pvqvr) n r]

(-pnq) v (pnr) v(qnr) v (rnr)

maybe?

Answer 28

that might be a cirle lol

Answer 29

aaaaaaaaaaaaaaaaaaaaaargh

Answer 30

\[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)\lor r\]
\[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land (\lnot r\lor r)\]
\[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land r\]
ugh is right

Answer 31

ok i think i may have it please check

Answer 32

starting here
\[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

Answer 33

-rvr = T right?

Answer 34

\[(\lnot p\land \lnot q)\lor r \lor (( p\lor q)\land \lnot r)\]
\[(\lnot p\land \lnot q)\lor (r\lor (p\lor q) \land (r \lor \lnot r)\]\]

Answer 35

\[(\lnot p\land \lnot q)\lor (r\lor (p\lor q)) \land T\]\]

Answer 36

\[((\lnot p \land \lnot q)\lor r \lor (p\lor q)\]

Answer 37

\[\lnot( p \lor q) \lor r \lor (p \lor q)\]
\[\lnot (p\lor q) \lor (p\lor q) \lor r)\]
\[T\lor r\]
TTTTTTTTTTTTTTTTTTTTTTTTTTT

Answer 38

would it be easier with truth table?

Answer 39

maybe? probably took 8 too many steps

Answer 40

looks good to me :)

Answer 41

i wanted to do it without truth tables and got all messed up

Answer 42

you could just do this with words

Answer 43

ok zarkon, lets see 2 steps

Answer 44

not is 2 steps

Answer 45

yeah i feel this way about all this nonsense
lots of symbols for the completely obvious

Answer 46

as a professor i once had (who sadly recently died)
"just because it is obvious, doesn't mean you can't prove it"

Answer 47

\[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\]

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

Answer 48

I should probably write ...this tells us p is false and q is false

Answer 49

[ (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)

Answer 50

no claims to being the shortest.

Answer 51

yes it looks nice and short!

Answer 52

sarcasm? :D

Answer 53

no no you should have see what i wrote!

Answer 54

im too tired to unravel it, beacuse you tend to go back, lol