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satellite73

  • 4 years ago

show \[[(p\rightarrow q)\land (q\rightarrow r)]\rightarrow (p\rightarrow r)\] is a tautology, without using truth tables

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  1. anonymous
    • 4 years ago
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    Seems like transitivity.

  2. amistre64
    • 4 years ago
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    -[(p>q)n(q>r)] v (p>r) -(p>q) v -(q>r) v (p>r) -(-pvq) v -(-qvr) v (-pvr) (pn-q) v (qn-r) v -p v r

  3. anonymous
    • 4 years ago
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    it is entirely obvious, but somehow i get messed up at the last step

  4. anonymous
    • 4 years ago
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    ok so i am fine at \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]

  5. amistre64
    • 4 years ago
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    (pn-q) v (qn-r) v-p vr -------------- (pv-p) n (-qv-p) v (qvr) n (-rvr) T n (-qv-p) v (qvr) n T (Tn-q) v (Tn-p) v (qnT) v (rnT) -q v-p v q v r (-qvq) v-p v r T v ..... = T

  6. anonymous
    • 4 years ago
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    Can't you just write that it represents a syllogism?

  7. anonymous
    • 4 years ago
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    @amistre, you lost me on the last line

  8. anonymous
    • 4 years ago
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    yeah i am trying to do it amistre way

  9. amistre64
    • 4 years ago
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    since we got all ors, and a T; the rest dont matter

  10. amistre64
    • 4 years ago
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    T or ? or ? or ? = T

  11. anonymous
    • 4 years ago
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    that is fine , but what is this (pn-q) v (qn-r) v-p vr ?

  12. amistre64
    • 4 years ago
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    oring -p to the first bit and r to the last bit i stacked them vertical; helps me see whats going on

  13. anonymous
    • 4 years ago
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    ok that is the line i need to get starting here \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]

  14. amistre64
    • 4 years ago
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    \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \lor r)\]

  15. amistre64
    • 4 years ago
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    a -> changes to "v" not "n"

  16. amistre64
    • 4 years ago
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    p -> r :: -p v r

  17. anonymous
    • 4 years ago
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    ah ok so we have \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \lor r)\] and then \[(p\land \lnot q)\lor (q\land \lnot r)\lor \lnot p \lor r\] and then we are going to commute to \[\lnot p \lor (p \land \lnot q) \lor (q \land \lnot r) \lor q\]

  18. amistre64
    • 4 years ago
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    yep

  19. amistre64
    • 4 years ago
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    not a vq, vr

  20. amistre64
    • 4 years ago
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    good, and from there its cake walk; hot coals cake walk, but cake walk nonetheless :)

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

  22. anonymous
    • 4 years ago
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    i think that is right now yes?

  23. amistre64
    • 4 years ago
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    yep; then we can T off the sides and distribute thru the middles which cancels out the ts

  24. anonymous
    • 4 years ago
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    god this makes my head spin. thanks

  25. amistre64
    • 4 years ago
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    yeah, i blame guass lol

  26. anonymous
    • 4 years ago
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    well i tried to do this on the fly and got hopelessly stuck, then i tried it again and got stuck again not on the fly. thanks for straightening it out!

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