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lgbasallote

  • 2 years ago

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

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  1. lgbasallote
    • 2 years ago
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    i suppse \[\neg p \wedge (p \vee q) \equiv p\]

  2. lgbasallote
    • 2 years ago
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    so then it becomes \[p \rightarrow q\]

  3. lgbasallote
    • 2 years ago
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    then what?

  4. lgbasallote
    • 2 years ago
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    oh i see

  5. lgbasallote
    • 2 years ago
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    i made a mistake

  6. lgbasallote
    • 2 years ago
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    \[\neg p \wedge (p \vee q) \equiv F \vee (\neg p \wedge q) \rightarrow q\]

  7. lgbasallote
    • 2 years ago
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    then this becomes \[T \wedge (p \vee q) \vee q\]

  8. lgbasallote
    • 2 years ago
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    then \[T \wedge (p \vee q) \equiv p \vee q \]

  9. lgbasallote
    • 2 years ago
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    in my solution i will be treating them the same....my latex is gonna get confusing if i follow the rules

  10. lgbasallote
    • 2 years ago
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    \[\equiv p \vee q \vee q\] \[q \vee q \equiv T\] so.. \[\equiv p \vee T \equiv T\] yes?

  11. lgbasallote
    • 2 years ago
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    so then this is a tautology?

  12. PhoenixFire
    • 2 years ago
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    implication law is P->Q=nP V Q Right? Sorry for no Latex,

  13. lgbasallote
    • 2 years ago
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    yes

  14. lgbasallote
    • 2 years ago
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    so is that a yes to my question?

  15. PhoenixFire
    • 2 years ago
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    yes it's a tautology, but you messed something up and somehow ended up with a tautology anyways lol

  16. lgbasallote
    • 2 years ago
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    where?

  17. UnkleRhaukus
    • 2 years ago
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    \[\small¬p∧(p∨q)\Rightarrow q\iff(\neg p\wedge p) \vee(\neg p\lor q)\Rightarrow q\iff (\neg p\lor q)\Rightarrow q \iff (\neg p\Rightarrow q)\vee(q\Rightarrow q )\]\[\iff q\Rightarrow q\qquad \top\]

  18. lgbasallote
    • 2 years ago
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    hmm looks different...

  19. lgbasallote
    • 2 years ago
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    implication is distributive?

  20. PhoenixFire
    • 2 years ago
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  21. lgbasallote
    • 2 years ago
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    seems you're the one who went wrong @PhoenixFire ....

  22. lgbasallote
    • 2 years ago
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    \[(\neg p \wedge q) \rightarrow q\] should become \[\neg(\neg p \wedge q) \vee q\] by DM law \[p \vee q \vee q\]

  23. lgbasallote
    • 2 years ago
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    hmm seems i missed a step too

  24. lgbasallote
    • 2 years ago
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    nevertheless, important thing is.. i was right....that was my question in the first place anyways

  25. PhoenixFire
    • 2 years ago
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    de morgans law.... you have to negate both. n(nP ^ Q) V Q becomes nnP V nQ V Q

  26. PhoenixFire
    • 2 years ago
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    nnP <-- Involution law becomes P. What you missed was distributing the negative to the Q during De Morgan's Law

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