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satellite73
 4 years ago
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\[[(p\rightarrow q)\land (q\rightarrow r)]\rightarrow (p\rightarrow r)\] is a tautology, without using truth tables
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Seems like transitivity.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5[(p>q)n(q>r)] v (p>r) (p>q) v (q>r) v (p>r) (pvq) v (qvr) v (pvr) (pnq) v (qnr) v p v r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is entirely obvious, but somehow i get messed up at the last step

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so i am fine at \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5(pnq) v (qnr) vp vr  (pvp) n (qvp) v (qvr) n (rvr) T n (qvp) v (qvr) n T (Tnq) v (Tnp) v (qnT) v (rnT) q vp v q v r (qvq) vp v r T v ..... = T

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can't you just write that it represents a syllogism?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre, you lost me on the last line

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah i am trying to do it amistre way

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5since we got all ors, and a T; the rest dont matter

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is fine , but what is this (pnq) v (qnr) vp vr ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5oring p to the first bit and r to the last bit i stacked them vertical; helps me see whats going on

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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
 4 years ago
Best ResponseYou've already chosen the best response.5\[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \lor r)\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5a > changes to "v" not "n"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah 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
 4 years ago
Best ResponseYou've already chosen the best response.5good, and from there its cake walk; hot coals cake walk, but cake walk nonetheless :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\lnot p \lor p )\land (\lnot p \lor \lnot q)\lor (q \lor r) \land (\lnot r \lor r)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think that is right now yes?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5yep; then we can T off the sides and distribute thru the middles which cancels out the ts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0god this makes my head spin. thanks

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.5yeah, i blame guass lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 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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