# Verify that $(\neg P\land N)\lor(\neg D \land P)\rightarrow \neg D\lor N$

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.

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

\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.

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