# Can a conclusion in syllogism be reversed?

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?

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

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$$\dots$$

• 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.
– liew
Commented Nov 15, 2019 at 7:21
• 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?
– liew
Commented Nov 22, 2019 at 0:13
• @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 $?$ Commented Nov 22, 2019 at 0:27
• 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
– liew
Commented Nov 22, 2019 at 0:39
• 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
– liew
Commented Nov 22, 2019 at 0:50