I think I understand how
(P Λ Q) ≡ ¬ (P ↓ Q) and
(P Λ Q) ≡ ¬P ↓ ¬Q.
However I can’t seem to make sense of
¬ (P ↓ Q) ≡ (P ↓ Q) ↓ (P ↓ Q) and
¬P ↓ ¬Q ≡ (P ↓ P) ↓ (Q ↓ Q).
Please help.
Thanks
The chief difficulty here is that almost no one agrees on notation.
Although it was invented in the 1880's by CS Pierce, still today there is a disagreement.
Here is the way I learned it from C.I. Copi. He called IT Stroke and dagger.
$\begin{array}{*{20}{c}}P&{}&Q&{}&{}&{P|Q}\\\hline T&{}&T&{}&{}&F\\T&{}&F&{}&{}&T\\F&{}&T&{}&{}&T\\F& {}&F&{}&{}&T\end{array}$ and $\begin{array}{*{20}{c}}P&{}&Q&{}&{}&{P \downarrow q} \\ \hline T&{}&T&{}&{}&F\\T&{}&F&{}&{}&F\\F&{}&T&{}&{}&F\\F& {}&F&{}&{}&T\end{array}$
Willard Quine (the greatest American logician of the 20th century) calls it alternative denial.
just methodically make the truth tables.
$\begin{array}{ccccc}
P &Q &(P \downarrow Q) &\neg (P \downarrow Q)& (P \downarrow Q) \downarrow(P \downarrow Q) \\
0 &0 &1 &0 &0 \\
0 &1 &0 &1 &1 \\
1 &0 &0 &1 &1 \\
1 &1 &0 &1 &1
\end{array}$
you can see the last two columns are identical.
$\begin{array}{ccddcccc}
P &Q &(P \downarrow P) &(Q\downarrow Q)& (P \downarrow P) \downarrow(Q\downarrow Q) &\neg P & \neg Q &\neg P \downarrow \neg Q \\
0 &0 &1 &1 &0 &1 &1 &0\\
0 &1 &1 &0 &0 &1 &0 &0\\
1 &0 &0 &1 &0 &0 &1 &0\\
1 &1 &0 &0 &1 &0 &0 &1\\
\end{array}$
you can see the 5th and 8th columns are identical
I’m sorry, I should have stated that I understood that they are equivalent and I could derive this via truth formulas. However although I know it’s correct I can’t understand the reasoning. The actual problem in its entirety was: Find formulas using only the connective ↓ that are equivalent to ¬P, P V Q and P Λ Q. I got that (P V Q) ≡ ¬ (P ↓ Q) [Sorry about the error as I originally posted this as (P Λ Q) ≡ ¬ (P ↓ Q)] The reasoning being that saying “either P OR Q is true” (P V Q), is the same as saying “it is not the case that neither P nor Q is true” ¬ (P ↓ Q). I must admit that I’m still haven’t completely wrapped my mind around this however how one gets from ¬ (P ↓ Q) to (P ↓ Q) ↓ (P ↓ Q) is to me completely unfathomable (reasoning wise that is). As ¬ (P ↓ Q) ≡ (P ↓ Q) ↓ (P ↓ Q). Hope I better explained myself.
So I take it that there is no way to reason ¬ (P ↓ Q) ≡ (P ↓ Q) ↓ (P ↓ Q) or ¬P ↓ ¬Q ≡ (P ↓ P) ↓ (Q ↓ Q). If so then thank you.
You see I recently started doing this and had the impression that it all could be somehow “reasoned” hence I thought I was missing something as I could not put into words how one got from ¬ (P ↓ Q) to (P ↓ Q) ↓ (P ↓ Q) or ¬P ↓ ¬Q to (P ↓ P) ↓ (Q ↓ Q). I have already derived it myself via tinkering with the truth tables it’s the rationalizing into words that I can’t get.
Thanks
As I wrote in my reply, the function is of great interest to professional logicians.
It turns out that all other logical functions can be written with the stroke,$\displaystyle \downarrow $, alone.
Quine called it the alternate denial, neither P nor Q is true.
So $\displaystyle \neg P \equiv \left( {P \downarrow P} \right)$ we have replaced not.
As for your questions above, those were just worked out as part of the replacement program.
If you have access to a good library Mathematical Logic by Quine, 1947 has a discussion.
But there is an advantage to expressing combinations of logic symbols in terms of others when taking into account the capabilities of digital logic circuits. For example, I believe all truth tables can be expressed using only “and” “or” and “not.”
If the dagger provides such an advantage, or potential advantage, that is certainly a reason.