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lgbasalloteBest ResponseYou've already chosen the best response.0
i suppse \[\neg p \wedge (p \vee q) \equiv p\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
so then it becomes \[p \rightarrow q\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
\[\neg p \wedge (p \vee q) \equiv F \vee (\neg p \wedge q) \rightarrow q\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
then this becomes \[T \wedge (p \vee q) \vee q\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
then \[T \wedge (p \vee q) \equiv p \vee q \]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
in my solution i will be treating them the same....my latex is gonna get confusing if i follow the rules
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
\[\equiv p \vee q \vee q\] \[q \vee q \equiv T\] so.. \[\equiv p \vee T \equiv T\] yes?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
so then this is a tautology?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
implication law is P>Q=nP V Q Right? Sorry for no Latex,
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
so is that a yes to my question?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
yes it's a tautology, but you messed something up and somehow ended up with a tautology anyways lol
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
\[\small¬p∧(p∨q)\Rightarrow q\iff(\neg p\wedge p) \vee(\neg p\lor q)\Rightarrow q\iff (\neg p\lor q)\Rightarrow q \iff (\neg p\Rightarrow q)\vee(q\Rightarrow q )\]\[\iff q\Rightarrow q\qquad \top\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
hmm looks different...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
implication is distributive?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
seems you're the one who went wrong @PhoenixFire ....
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
\[(\neg p \wedge q) \rightarrow q\] should become \[\neg(\neg p \wedge q) \vee q\] by DM law \[p \vee q \vee q\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
hmm seems i missed a step too
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
nevertheless, important thing is.. i was right....that was my question in the first place anyways
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
de morgans law.... you have to negate both. n(nP ^ Q) V Q becomes nnP V nQ V Q
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.0
nnP < Involution law becomes P. What you missed was distributing the negative to the Q during De Morgan's Law
 one year ago
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