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anonymous
 3 years ago
Evaluate: \[[\neg p \wedge (p \vee q)] \rightarrow q\]
anonymous
 3 years ago
Evaluate: \[[\neg p \wedge (p \vee q)] \rightarrow q\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i suppse \[\neg p \wedge (p \vee q) \equiv p\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so then it becomes \[p \rightarrow q\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\neg p \wedge (p \vee q) \equiv F \vee (\neg p \wedge q) \rightarrow q\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then this becomes \[T \wedge (p \vee q) \vee q\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then \[T \wedge (p \vee q) \equiv p \vee q \]

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

anonymous
 3 years ago
Best 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so then this is a tautology?

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.0implication law is P>Q=nP V Q Right? Sorry for no Latex,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so is that a yes to my question?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm looks different...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0implication is distributive?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0seems you're the one who went wrong @PhoenixFire ....

anonymous
 3 years ago
Best 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 \[p \vee q \vee q\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm seems i missed a step too

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nevertheless, important thing is.. i was right....that was my question in the first place anyways

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

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