satellite73
  • satellite73
Fewest number of steps, no truth tables show \[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] is a tautology
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
fewest steps would be the wolf I think
anonymous
  • anonymous
i know it is true and i can see why, but i cannot make the steps
amistre64
  • amistre64
p>r = -pvr q>r = -qvr

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anonymous
  • anonymous
i don't want a truth table
amistre64
  • amistre64
these are identities and operations
amistre64
  • amistre64
[]>r = -[]vr
anonymous
  • anonymous
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 \[\lnot[(p\lor q)\land(\lnot p\lor r)\land (\lnot q \lor r)]\lor r\]
amistre64
  • amistre64
-(pvq)-n-(-pvr)-n-(-qvr) v r (-pn-q) v (pn-r) v (qn-r) v r
anonymous
  • anonymous
\[[\lnot(p\lor q)\lor \lnot(\lnot p\lor r)\lor \lnot(\lnot q \lor r)]\lor r\]
amistre64
  • amistre64
(-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
anonymous
  • anonymous
ok just about where you are \[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\]
amistre64
  • amistre64
(-pn-q) v r v q (-pv q) n (-qv q) v r =q (-p v q v r ) n q = q perhaps
anonymous
  • anonymous
problem is i want T
amistre64
  • amistre64
thats not a T tho is it
anonymous
  • anonymous
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
amistre64
  • amistre64
undistribute maybe
anonymous
  • anonymous
oh maybe i can factor a not r out of the last two terms before the "or r"??
amistre64
  • amistre64
yep
anonymous
  • anonymous
which gets me what?
amistre64
  • amistre64
(pvq) n -r
anonymous
  • anonymous
maybe \[(p\lor q)\land \lnot r\]
amistre64
  • amistre64
since you got all vs inbetween the [ ] seem rather useless
anonymous
  • anonymous
ok so \[[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)]\lor r\] \[[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]
amistre64
  • amistre64
distribute your vr threw
anonymous
  • anonymous
\[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]
anonymous
  • anonymous
looks like i should be close right??
amistre64
  • amistre64
perhaps (-pnq) v [(pvqvr) n (-rvr)] =r (-pnq) v [(pvqvr) n r] (-pnq) v (pnr) v(qnr) v (rnr) maybe?
amistre64
  • amistre64
that might be a cirle lol
anonymous
  • anonymous
aaaaaaaaaaaaaaaaaaaaaargh
amistre64
  • amistre64
\[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)\lor r\] \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land (\lnot r\lor r)\] \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land r\] ugh is right
anonymous
  • anonymous
ok i think i may have it please check
anonymous
  • anonymous
starting here \[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]
amistre64
  • amistre64
-rvr = T right?
anonymous
  • anonymous
\[(\lnot p\land \lnot q)\lor r \lor (( p\lor q)\land \lnot r)\] \[(\lnot p\land \lnot q)\lor (r\lor (p\lor q) \land (r \lor \lnot r)\]\]
anonymous
  • anonymous
\[(\lnot p\land \lnot q)\lor (r\lor (p\lor q)) \land T\]\]
anonymous
  • anonymous
\[((\lnot p \land \lnot q)\lor r \lor (p\lor q)\]
anonymous
  • anonymous
\[\lnot( p \lor q) \lor r \lor (p \lor q)\] \[\lnot (p\lor q) \lor (p\lor q) \lor r)\] \[T\lor r\] TTTTTTTTTTTTTTTTTTTTTTTTTTT
anonymous
  • anonymous
would it be easier with truth table?
anonymous
  • anonymous
maybe? probably took 8 too many steps
amistre64
  • amistre64
looks good to me :)
anonymous
  • anonymous
i wanted to do it without truth tables and got all messed up
Zarkon
  • Zarkon
you could just do this with words
anonymous
  • anonymous
ok zarkon, lets see 2 steps
Zarkon
  • Zarkon
not is 2 steps
anonymous
  • anonymous
yeah i feel this way about all this nonsense lots of symbols for the completely obvious
anonymous
  • anonymous
as a professor i once had (who sadly recently died) "just because it is obvious, doesn't mean you can't prove it"
Zarkon
  • Zarkon
\[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] 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
  • Zarkon
I should probably write ...this tells us p is false and q is false
phi
  • phi
[ (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
  • phi
no claims to being the shortest.
anonymous
  • anonymous
yes it looks nice and short!
anonymous
  • anonymous
sarcasm? :D
anonymous
  • anonymous
no no you should have see what i wrote!
anonymous
  • anonymous
im too tired to unravel it, beacuse you tend to go back, lol

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