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

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1. anonymous

i suppse $\neg p \wedge (p \vee q) \equiv p$

2. anonymous

so then it becomes $p \rightarrow q$

3. anonymous

then what?

4. anonymous

oh i see

5. anonymous

i made a mistake

6. anonymous

$\neg p \wedge (p \vee q) \equiv F \vee (\neg p \wedge q) \rightarrow q$

7. anonymous

then this becomes $T \wedge (p \vee q) \vee q$

8. anonymous

then $T \wedge (p \vee q) \equiv p \vee q$

9. anonymous

in my solution i will be treating them the same....my latex is gonna get confusing if i follow the rules

10. anonymous

$\equiv p \vee q \vee q$ $q \vee q \equiv T$ so.. $\equiv p \vee T \equiv T$ yes?

11. anonymous

so then this is a tautology?

12. PhoenixFire

implication law is P->Q=nP V Q Right? Sorry for no Latex,

13. anonymous

yes

14. anonymous

so is that a yes to my question?

15. PhoenixFire

yes it's a tautology, but you messed something up and somehow ended up with a tautology anyways lol

16. anonymous

where?

17. UnkleRhaukus

$\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$

18. anonymous

hmm looks different...

19. anonymous

implication is distributive?

20. PhoenixFire

21. anonymous

seems you're the one who went wrong @PhoenixFire ....

22. anonymous

$(\neg p \wedge q) \rightarrow q$ should become $\neg(\neg p \wedge q) \vee q$ by DM law $p \vee q \vee q$

23. anonymous

hmm seems i missed a step too

24. anonymous

nevertheless, important thing is.. i was right....that was my question in the first place anyways

25. PhoenixFire

de morgans law.... you have to negate both. n(nP ^ Q) V Q becomes nnP V nQ V Q

26. PhoenixFire

nnP <-- Involution law becomes P. What you missed was distributing the negative to the Q during De Morgan's Law

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