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.

  • $\begingroup$ The third $\equiv$ is only a $\to$... $\endgroup$ Jan 20, 2022 at 12:28
  • $\begingroup$ Did you checked it with truth table? $\endgroup$ Jan 20, 2022 at 12:31
  • $\begingroup$ @MauroALLEGRANZA is it right after edited? this is draw from math.stackexchange.com/questions/4361214/… $\endgroup$
    – LJNG
    Jan 20, 2022 at 13:00
  • $\begingroup$ @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 $\endgroup$
    – LJNG
    Jan 20, 2022 at 13:34
  • 1
    $\begingroup$ @LJNG The third inference… the implication. (It would be valid in the opposite direction that you used it.) $\endgroup$ Jan 20, 2022 at 19:56

2 Answers 2


\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.

  • $\begingroup$ 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.) $\endgroup$ Feb 7, 2022 at 16:58
  • $\begingroup$ I fixed it now. $\endgroup$ Feb 7, 2022 at 17:04
  • $\begingroup$ @spaceisdarkgreen Ah, haha, thanks for the observation and fixing the Question post; I've edited my answer, in response. $\endgroup$
    – ryang
    Feb 7, 2022 at 17:22

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.


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