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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

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  1. amistre64
    • 4 years ago
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    fewest steps would be the wolf I think

  2. anonymous
    • 4 years ago
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    i know it is true and i can see why, but i cannot make the steps

  3. amistre64
    • 4 years ago
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    p>r = -pvr q>r = -qvr

  4. anonymous
    • 4 years ago
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    i don't want a truth table

  5. amistre64
    • 4 years ago
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    these are identities and operations

  6. amistre64
    • 4 years ago
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    []>r = -[]vr

  7. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    -(pvq)-n-(-pvr)-n-(-qvr) v r (-pn-q) v (pn-r) v (qn-r) v r

  9. anonymous
    • 4 years ago
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    \[[\lnot(p\lor q)\lor \lnot(\lnot p\lor r)\lor \lnot(\lnot q \lor r)]\lor r\]

  10. amistre64
    • 4 years ago
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    (-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
    • 4 years ago
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    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
    • 4 years ago
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    (-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
    • 4 years ago
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    problem is i want T

  14. amistre64
    • 4 years ago
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    thats not a T tho is it

  15. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    undistribute maybe

  17. anonymous
    • 4 years ago
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    oh maybe i can factor a not r out of the last two terms before the "or r"??

  18. amistre64
    • 4 years ago
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    yep

  19. anonymous
    • 4 years ago
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    which gets me what?

  20. amistre64
    • 4 years ago
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    (pvq) n -r

  21. anonymous
    • 4 years ago
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    maybe \[(p\lor q)\land \lnot r\]

  22. amistre64
    • 4 years ago
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    since you got all vs inbetween the [ ] seem rather useless

  23. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    distribute your vr threw

  25. anonymous
    • 4 years ago
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    \[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

  26. anonymous
    • 4 years ago
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    looks like i should be close right??

  27. amistre64
    • 4 years ago
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    perhaps (-pnq) v [(pvqvr) n (-rvr)] =r (-pnq) v [(pvqvr) n r] (-pnq) v (pnr) v(qnr) v (rnr) maybe?

  28. amistre64
    • 4 years ago
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    that might be a cirle lol

  29. anonymous
    • 4 years ago
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    aaaaaaaaaaaaaaaaaaaaaargh

  30. amistre64
    • 4 years ago
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    \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\land \lnot r)\lor r\] \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land (\lnot r\lor r)\] \[(\lnot p\land \lnot q)\lor ( p\land \lnot r)\lor (q\lor r) \land r\] ugh is right

  31. anonymous
    • 4 years ago
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    ok i think i may have it please check

  32. anonymous
    • 4 years ago
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    starting here \[(\lnot p\land \lnot q)\lor (( p\lor q)\land \lnot r)\lor r\]

  33. amistre64
    • 4 years ago
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    -rvr = T right?

  34. anonymous
    • 4 years ago
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    \[(\lnot p\land \lnot q)\lor r \lor (( p\lor q)\land \lnot r)\] \[(\lnot p\land \lnot q)\lor (r\lor (p\lor q) \land (r \lor \lnot r)\]\]

  35. anonymous
    • 4 years ago
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    \[(\lnot p\land \lnot q)\lor (r\lor (p\lor q)) \land T\]\]

  36. anonymous
    • 4 years ago
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    \[((\lnot p \land \lnot q)\lor r \lor (p\lor q)\]

  37. anonymous
    • 4 years ago
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    \[\lnot( p \lor q) \lor r \lor (p \lor q)\] \[\lnot (p\lor q) \lor (p\lor q) \lor r)\] \[T\lor r\] TTTTTTTTTTTTTTTTTTTTTTTTTTT

  38. anonymous
    • 4 years ago
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    would it be easier with truth table?

  39. anonymous
    • 4 years ago
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    maybe? probably took 8 too many steps

  40. amistre64
    • 4 years ago
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    looks good to me :)

  41. anonymous
    • 4 years ago
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    i wanted to do it without truth tables and got all messed up

  42. Zarkon
    • 4 years ago
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    you could just do this with words

  43. anonymous
    • 4 years ago
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    ok zarkon, lets see 2 steps

  44. Zarkon
    • 4 years ago
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    not is 2 steps

  45. anonymous
    • 4 years ago
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    yeah i feel this way about all this nonsense lots of symbols for the completely obvious

  46. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    \[[(p\lor q)\land (p\rightarrow r)\land (q\rightarrow r)]\rightarrow r\] 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
    • 4 years ago
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    I should probably write ...this tells us p is false and q is false

  49. phi
    • 4 years ago
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    [ (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
    • 4 years ago
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    no claims to being the shortest.

  51. anonymous
    • 4 years ago
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    yes it looks nice and short!

  52. anonymous
    • 4 years ago
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    sarcasm? :D

  53. anonymous
    • 4 years ago
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    no no you should have see what i wrote!

  54. anonymous
    • 4 years ago
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    im too tired to unravel it, beacuse you tend to go back, lol

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