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Evaluate: \[[\neg p \wedge (p \vee q)] \rightarrow q\]

Discrete Math
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i suppse \[\neg p \wedge (p \vee q) \equiv p\]
so then it becomes \[p \rightarrow q\]
then what?

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Other answers:

oh i see
i made a mistake
\[\neg p \wedge (p \vee q) \equiv F \vee (\neg p \wedge q) \rightarrow q\]
then this becomes \[T \wedge (p \vee q) \vee q\]
then \[T \wedge (p \vee q) \equiv p \vee q \]
in my solution i will be treating them the same....my latex is gonna get confusing if i follow the rules
\[\equiv p \vee q \vee q\] \[q \vee q \equiv T\] so.. \[\equiv p \vee T \equiv T\] yes?
so then this is a tautology?
implication law is P->Q=nP V Q Right? Sorry for no Latex,
yes
so is that a yes to my question?
yes it's a tautology, but you messed something up and somehow ended up with a tautology anyways lol
where?
\[\small¬p∧(p∨q)\Rightarrow q\iff(\neg p\wedge p) \vee(\neg p\lor q)\Rightarrow q\iff (\neg p\lor q)\Rightarrow q \iff (\neg p\Rightarrow q)\vee(q\Rightarrow q )\]\[\iff q\Rightarrow q\qquad \top\]
hmm looks different...
implication is distributive?
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seems you're the one who went wrong @PhoenixFire ....
\[(\neg p \wedge q) \rightarrow q\] should become \[\neg(\neg p \wedge q) \vee q\] by DM law \[p \vee q \vee q\]
hmm seems i missed a step too
nevertheless, important thing is.. i was right....that was my question in the first place anyways
de morgans law.... you have to negate both. n(nP ^ Q) V Q becomes nnP V nQ V Q
nnP <-- Involution law becomes P. What you missed was distributing the negative to the Q during De Morgan's Law

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