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The answer seems to be $\neg D \lor N$, but below is what I got,no idea where it goes wrong

$$\begin{aligned} (\neg P\land N)\lor(\neg D \land P) &\equiv \neg(\neg N\lor P)\lor \neg(\neg P\lor D)\\ &\equiv\neg[(N\implies P)\land(P\implies D)]\\&\rightarrow \neg(N\implies D)\\ &\equiv\neg(\neg N\lor D) \\ &\equiv\neg D\land N \end{aligned} $$

Appreciate for any help.

spaceisdarkgreen
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LJNG
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  • The third $\equiv$ is only a $\to$... – Mauro ALLEGRANZA Jan 20 '22 at 12:28
  • Did you checked it with truth table? – Mauro ALLEGRANZA Jan 20 '22 at 12:31
  • @MauroALLEGRANZA is it right after edited? this is draw from https://math.stackexchange.com/questions/4361214/how-to-show-negd-land-p-land-n-lor-p-implies-neg-d-lor-n-by-propositio – LJNG Jan 20 '22 at 13:00
  • @MauroALLEGRANZA Thank you for your reply. What he said is $\neg D\lor N$, but what I got is $\neg D\land N,$ I have no idea where it goes wrong in my algebraic operation – LJNG Jan 20 '22 at 13:34
  • Please, check it with truth assignment: $v(N)=T$ and $v(D)=T$. – Mauro ALLEGRANZA Jan 20 '22 at 13:40
  • But the ERROR is the same... – Mauro ALLEGRANZA Jan 20 '22 at 13:47
  • I have no idea about what you said. much appreciate if you can let me know what my process is wrong that I get $\neg D\land N$ – LJNG Jan 20 '22 at 13:50
  • @MauroALLEGRANZA thank you for your reply. I've never seen this inference, but this seems to analogise to all elements in two sets intersection must also in their union – LJNG Jan 20 '22 at 14:05
  • @LJNG No it’s still not right. The implication to $\lnot D\land N$ is invalid. If N is true, P is false and D is true, the premise is true and the conclusion is false. – spaceisdarkgreen Jan 20 '22 at 18:12
  • @spaceisdarkgreen thank you for your reply. where is my deduce process gone wrong? – LJNG Jan 20 '22 at 19:03
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    @LJNG The third inference… the implication. (It would be valid in the opposite direction that you used it.) – spaceisdarkgreen Jan 20 '22 at 19:56
  • @spaceisdarkgreen I don't think I get you, would you please leave your answer rather than comments. Cheers – LJNG Jan 20 '22 at 21:18
  • @LJNG No, I only have a phone right now and don’t have time to type it out. My advice would be to take the truth values for N, D, and P I mentioned above and check at which step the expression’s truth value switches from true to false. – spaceisdarkgreen Jan 20 '22 at 23:21

2 Answers2

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\begin{aligned} (\neg P\land N)\lor(\neg D \land P) &\equiv \neg(\neg N\lor P)\lor \neg(\neg P\lor D)\\ &\equiv\neg((N\implies P)\land(P\implies D))\\&\rightarrow \neg(N\implies D)\\ &\equiv\neg(\neg N\lor D) \\ &\equiv\neg D\land N \end{aligned}

That logical entailment in Line 3 is incorrect, as evidenced by the assignment $(P,N,D)=(0,1,1).$

Here's a correct attempt, if you don't mind applying the distributive laws: \begin{aligned} (\neg P\land N)\lor(\neg D \land P) &\equiv (¬P∨¬D)∧(¬P∨P)∧(N∨¬D)∧(N∨P)\\ &\models (N∨¬D)\\ &\equiv \neg D ∨ N. \end{aligned} Thus, $$(\neg P\land N)\lor(\neg D \land P) \to \neg D ∨ N$$ is a validity, as required.

P.S. I use $≡$ and $⊨$ to mean logically equivalent and logically implies, respectively (i.e., as metalogical symbols), while I use $\to$ merely as the material conditional (i.e., as a logical operator). As for $\implies,$ I use it just to mean implies (e.g., $x=2\implies x^2=4$) rather than as the material conditional.

ryang
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  • I think you misunderstand their attempt (the third line’s indentation is misleading… it is supposed to be a new line which is implied by the previous line, not a continuation of the sentence on the second line.) (Now I see it’s actually explicitly that way now, since someone edited it with the same reading you had.) – spaceisdarkgreen Feb 07 '22 at 16:58
  • I fixed it now. – spaceisdarkgreen Feb 07 '22 at 17:04
  • @spaceisdarkgreen Ah, haha, thanks for the observation and fixing the Question post; I've edited my answer, in response. – ryang Feb 07 '22 at 17:22
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It's easy to verify the relationship using Natural Deduction:

$ \dfrac { ((\neg p\land n) \vee (\neg d \land p)) \quad \dfrac{\dfrac{[(\neg p \land n)]}{n}(\land E)}{(\neg d \lor n)}(\lor I) \quad \dfrac{\dfrac{[(\neg d \land p)]}{(\neg d)}(\land E)}{(\neg d \lor n)}(\lor I) } { (\neg d \lor n) }(\lor E) $

PS: This is my second answer. The first one, as pointed out by
spaceisdarkgreen, was completely wrong.

lcmpereira
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