satellite73
  • satellite73
show \[[(p\rightarrow q)\land (q\rightarrow r)]\rightarrow (p\rightarrow r)\] is a tautology, without using truth tables
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Seems like transitivity.
amistre64
  • amistre64
-[(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
anonymous
  • anonymous
it is entirely obvious, but somehow i get messed up at the last step

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anonymous
  • anonymous
ok so i am fine at \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]
amistre64
  • amistre64
(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
anonymous
  • anonymous
Can't you just write that it represents a syllogism?
anonymous
  • anonymous
@amistre, you lost me on the last line
anonymous
  • anonymous
yeah i am trying to do it amistre way
amistre64
  • amistre64
since we got all ors, and a T; the rest dont matter
amistre64
  • amistre64
T or ? or ? or ? = T
anonymous
  • anonymous
that is fine , but what is this (pn-q) v (qn-r) v-p vr ?
amistre64
  • amistre64
oring -p to the first bit and r to the last bit i stacked them vertical; helps me see whats going on
anonymous
  • anonymous
ok that is the line i need to get starting here \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]
amistre64
  • amistre64
\[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \lor r)\]
amistre64
  • amistre64
a -> changes to "v" not "n"
amistre64
  • amistre64
p -> r :: -p v r
anonymous
  • anonymous
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 \[\lnot p \lor (p \land \lnot q) \lor (q \land \lnot r) \lor q\]
amistre64
  • amistre64
yep
amistre64
  • amistre64
not a vq, vr
amistre64
  • amistre64
good, and from there its cake walk; hot coals cake walk, but cake walk nonetheless :)
anonymous
  • anonymous
\[(\lnot p \lor p )\land (\lnot p \lor \lnot q)\lor (q \lor r) \land (\lnot r \lor r)\]
anonymous
  • anonymous
i think that is right now yes?
amistre64
  • amistre64
yep; then we can T off the sides and distribute thru the middles which cancels out the ts
anonymous
  • anonymous
god this makes my head spin. thanks
amistre64
  • amistre64
yeah, i blame guass lol
anonymous
  • anonymous
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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