## 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

1. amistre64

fewest steps would be the wolf I think

2. anonymous

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

3. amistre64

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

4. anonymous

i don't want a truth table

5. amistre64

these are identities and operations

6. amistre64

[]>r = -[]vr

7. 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 $\lnot[(p\lor q)\land(\lnot p\lor r)\land (\lnot q \lor r)]\lor r$

8. amistre64

-(pvq)-n-(-pvr)-n-(-qvr) v r (-pn-q) v (pn-r) v (qn-r) v r

9. anonymous

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

10. 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

11. anonymous

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

12. amistre64

(-pn-q) v r v q (-pv q) n (-qv q) v r =q (-p v q v r ) n q = q perhaps

13. anonymous

problem is i want T

14. amistre64

thats not a T tho is it

15. 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

16. amistre64

undistribute maybe

17. anonymous

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

18. amistre64

yep

19. anonymous

which gets me what?

20. amistre64

(pvq) n -r

21. anonymous

maybe $(p\lor q)\land \lnot r$

22. amistre64

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

23. anonymous

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

24. amistre64

25. anonymous

$(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r$

26. anonymous

looks like i should be close right??

27. amistre64

perhaps (-pnq) v [(pvqvr) n (-rvr)] =r (-pnq) v [(pvqvr) n r] (-pnq) v (pnr) v(qnr) v (rnr) maybe?

28. amistre64

that might be a cirle lol

29. anonymous

aaaaaaaaaaaaaaaaaaaaaargh

30. 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

31. anonymous

ok i think i may have it please check

32. anonymous

starting here $(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r$

33. amistre64

-rvr = T right?

34. anonymous

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

35. anonymous

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

36. anonymous

$((\lnot p \land \lnot q)\lor r \lor (p\lor q)$

37. 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

38. anonymous

would it be easier with truth table?

39. anonymous

maybe? probably took 8 too many steps

40. amistre64

looks good to me :)

41. anonymous

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

42. Zarkon

you could just do this with words

43. anonymous

ok zarkon, lets see 2 steps

44. Zarkon

not is 2 steps

45. anonymous

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

46. anonymous

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

47. 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

48. Zarkon

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

49. 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)

50. phi

no claims to being the shortest.

51. anonymous

yes it looks nice and short!

52. anonymous

sarcasm? :D

53. anonymous

no no you should have see what i wrote!

54. anonymous

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