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Let's take Barbara:

M a P     All M are P
S a M     and all S are M
_____
S a P     thus: all S are P

It seems to work just fine if we reverse the conclusion:

M a P     All M are P
S a M     and all S are M
_____
P a S     thus all P are S

But what about other types? Would that always work? Let's take a look at Camestros and Camestros with reversed conclusion:

P a M     All P are M
S e M     and no S is M
_____
S o P     thus some S are not P

vs

P a M     All P are M
S e M     and no S is M
_____
P o S     thus some P are not S

Interpreting this type using Venn diagrams reveals that this one also works. I couldn't find any counter-example but didn't go through all 24 valid types. Wanted to ask - has anyone tried checking that?

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  • $\begingroup$ you can verify it in first order logic, which will make it eaiser $\endgroup$
    – Ethan
    Commented Nov 15, 2019 at 6:07
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    $\begingroup$ the first reversed statement is not true, but the second is true $\endgroup$
    – Ethan
    Commented Nov 15, 2019 at 6:30

1 Answer 1

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If we write them in FOL, it will be eaiser to check:

$\text{AAA(1)}:(\forall x,M(x)\rightarrow P(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\forall x,S(x)\rightarrow P(x))\tag{1}$

$\text{EAE(1)}:(\forall x,M(x)\rightarrow\neg P(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\forall x,S(x)\rightarrow\neg P(x))$ $\text{AII(1)}:(\forall x,M(x)\rightarrow P(x))\wedge(\exists x,S(x)\wedge M(x))\rightarrow(\exists x,S(x)\wedge P(x))$ $\text{EIO(1)}:(\forall x,M(x)\rightarrow\neg P(x))\wedge(\exists x,S(x)\wedge M(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AAI(1)}:(\forall x,M(x)\rightarrow P(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\exists x,S(x)\wedge P(x))$ $\text{EAO(1)}:(\forall x,M(x)\rightarrow\neg P(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AEE(2)}:(\forall x,P(x)\rightarrow M(x))\wedge(\forall x,S(x)\rightarrow\neg M(x))\rightarrow(\forall x,S(x)\rightarrow\neg P(x))$

$\text{EAE(2)}:(\forall x,P(x)\rightarrow\neg M(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\forall x,S(x)\rightarrow\neg P(x))\tag{2}$

$\text{AOO(2)}:(\forall x,P(x)\rightarrow M(x))\wedge(\exists x,S(x)\wedge\neg M(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{EIO(2)}:(\forall x,P(x)\rightarrow\neg M(x))\wedge(\exists x,S(x)\wedge M(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AEO(2)}:(\forall x,P(x)\rightarrow M(x))\wedge(\forall x,S(x)\rightarrow\neg M(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{EAO(2)}:(\forall x,P(x)\rightarrow\neg M(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AII(3)}:(\forall x,M(x)\rightarrow P(x))\wedge(\exists x,M(x)\wedge S(x))\rightarrow(\exists x,S(x)\wedge P(x))$ $\text{IAI(3)}:(\exists x,M(x)\wedge P(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge P(x))$ $\text{OAO(3)}:(\exists x,M(x)\wedge\neg P(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{EIO(3)}:(\forall x,M(x)\rightarrow\neg P(x))\wedge(\exists x,M(x)\wedge S(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AAI(3)}:(\forall x,M(x)\rightarrow P(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge P(x))$ $\text{EAO(3)}:(\forall x,M(x)\rightarrow\neg P(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AEE(4)}:(\forall x,P(x)\rightarrow M(x))\wedge(\forall x,M(x)\rightarrow\neg S(x))\rightarrow(\forall x,S(x)\rightarrow\neg P(x))$ $\text{IAI(4)}:(\exists x,P(x)\wedge M(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge P(x))$ $\text{EIO(4)}:(\forall x,P(x)\rightarrow\neg M(x))\wedge(\exists x,M(x)\wedge S(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AEO(4)}:(\forall x,P(x)\rightarrow M(x))\wedge(\forall x,M(x)\rightarrow\neg S(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{EAO(4)}:(\forall x,P(x)\rightarrow\neg M(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge\neg P(x))$ $\text{AAI(4)}:(\forall x,P(x)\rightarrow M(x))\wedge(\forall x,M(x)\rightarrow S(x))\rightarrow(\exists x,S(x)\wedge P(x))$


Here is the first statement with a reversed conclusion:

$$(\forall x,M(x)\rightarrow P(x))\wedge(\forall x,S(x)\rightarrow M(x))\rightarrow(\forall x,P(x)\rightarrow S(x))\tag*{*AAA(1)}$$

This will not always hold, for example, all Males are Poor and all Saltfish are Male, but we can not say all Poor things are Saltfish, take the world with a Male that is Poor but not a Saltfish, that (is Male$\to$ is Poor) hold also since it's not a Saltfish, second condition is vacuous true. but we can not conclude that he is a Saltfish from those condition, and actually he is not, therefore the statement is not true.

Second is true, since $S(x)\to M(x)$ and the contrapositive of $P(x)\to\neg M(x)$ is $M(x)\to\neg P(x)$ that we can prove $S(x)\to\neg P(x)$.

Similarly, we can check this for the rest of the statements$\dots$

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    $\begingroup$ Yeah, you're right, FOL approach seems like a smart idea since the process of checking could be speeded up by using some tautology checking software. I've found these diagrams for all valid types though: en.wikipedia.org/wiki/Syllogism#/media/File:Modus_Barbara.svg Looking at this also makes the fact that there are P that are not S obvious. Thanks a lot for answering the question and pointing out that mistake. $\endgroup$
    – liew
    Commented Nov 15, 2019 at 7:21
  • $\begingroup$ One more thing bothers me - why do we say that there are 24 valid types if we can actually provide more of them by using this trick? $\endgroup$
    – liew
    Commented Nov 22, 2019 at 0:13
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    $\begingroup$ @liew I think it's about AEIO and $4$ figures, can you find some examples that is valid and use AEIO statements in that $4$ figures but not in one of the $24$ types $?$ $\endgroup$
    – Ethan
    Commented Nov 22, 2019 at 0:27
  • $\begingroup$ Yeah, you're probably right. So it's all about premises - you can create only 24 schemes that produce valid reasoning, nobody cared about adding reversed conclusions when the list was created since that didn't matter that much - as you noticed figures are desribed only by premises layout $\endgroup$
    – liew
    Commented Nov 22, 2019 at 0:39
  • $\begingroup$ Btw it seems that conclusions in all types except for Barbara can be reversed without harm, maybe that's another reason why nobody cared about that $\endgroup$
    – liew
    Commented Nov 22, 2019 at 0:50

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