# Some questions regarding the proof of validity and invalidity of arguments in Logic.

I recently started reading Logic and read the 1st order logic or predicate logic, where the main problem is to prove the validity or invalidity of an argument.

While proving validity or invalidity it's the first task to have an intuition about the validity or invalidity of the argument,then you can proceed to solve the problem.My question is about that intuition.

e.g consider the problem :

1. All tenors are either overweight or effeminate.No overweight tenor is effeminate.Some tenors are effeminate.Therefore some tenors are overweight.

In this problem I thought in such a way : as all tenors are either overweight or effeminate,and as some tenors are effeminate,so all tenors are effeminate,hence the argument is invalid and I looked for a substitution instance in which the conclusion is false while the premises are true,and fortunately I found and the answer was right.

But in the next problem viz.,

• All tenors are either overweight or effeminate.No overweight tenor is effeminate.Some tenors are effeminate.Therefore some tenors are not overweight. I thought in the same way and found that as some tenors are effeminate so there is some tenors which are not overweight as all tenors are either overweight or effeminate.So it is a valid argument but unfortunately it was not true $($ I think as I found a substitution instance in which the conclusion is false but the premises are true. If we write down the problem in symbolic notation then we have : The premises are

($\forall x)(Tx \rightarrow (Ox \lor Ex))$

($\forall x)((Tx \land Ox) \rightarrow \sim Ex)$

($\exists x)(Tx \land Ex)$.

The conclusion is ($\exists x)(Tx \land \sim Ox)$

Now consider a universe consisting of ${x,y,}$ such that $Tx ,Ty , Ey$ are true and others are false.

Then the conclusion becomes false,still the premises are true.Hence the argument is invalid. $)$

So what's wrong in my intuition ? Can anyone please help me to point it out ? Also is there another way to make good intuition in such problems,or only experience can led to right intuition ?

Thanks.

I don't understand your proposed counterexample. Which of $x$ and $y$ are overweight? When you write

and others are false,

that suggests to me that neither $x$ nor $y$ satisfy $O$; but then clearly the conclusion $\exists x(Tx\wedge\neg Ox)$ is true there (both $x$ and $y$ satisfy $Tx\wedge\neg Ox$). (And incidentally, one of the axioms - $\forall x(Tx\implies (Ox\vee Ex))$ - is false, since neither $Ox$ nor $Ex$ hold of $y$.)

So maybe this isn't what you are trying to say in your proposed counterexample; in that case, what is your proposed counterexample? Which of $x$ and $y$ satisfy $O$? Which of $x$ and $y$ satisfy $E$? (In fact, no counterexample exists - the conclusion is indeed always true whenever the axioms are true.)

• Sorry ! I didn't get you . Will you explain ? Aug 23 '17 at 3:50
• @HirenGarai Can you clarify your proposed counterexample - is $Ox$ true? What about $Oy$? Aug 23 '17 at 3:51
• @HirenGarai In your proposed counterexample you have $Tx$, but neither $Ex$ nor $Ox$, and hence axiom 1 is false, rather than true. Aug 23 '17 at 3:51
• (@Bram28 And in fact the conclusion is true, too - so this fails to be a counterexample for two reasons!) Aug 23 '17 at 3:51
• @NoahSchweber Quite so! Aug 23 '17 at 3:52

Your reasoning in the first one isn't correct - you're right that it's invalid, but not why. You said your reasoning was that since all tenors are overweight or effeminate, and some are effeminate, they all must be effeminate - this is like saying that since all the chess pieces are white or black and some are white, they all are white.

As Noah pointed out in his answer, your counterexample isn't a counterexample. It looks like your approach to the intuition behind it isn't quite working out right. You seem like you're just trying to get down to the logic straight away - use everyday common sense instead for this part. For example, in the first situation, none of the premises "make" overweight tenors - the first premise allows them to be effeminate instead, the second premise describes tenors that are already overweight, and the third premise doesn't refer to overweight tenors at all. Nothing like "all X's are overweight tenors" shows up. That, to me, suggests there's something wrong with the argument - there seems to be something missing. Based on that intuition, I then look at an "edge case"; can we break the argument by making something extreme happen. In this case, to refute the conclusion we'd just need to know that no tenors are overweight. Is that possible? Well, to make the first premise true we'd just need all tenors to be effeminate. Then the second premise is vacuously true (it talks about overweight tenors, and there aren't any) and the third premise is true as well. So in a universe in which every tenor is effeminate, we have all three premises true and the conclusion false.

In your second problem, we see the opposite phenomenon - the conclusion is talking about non-overweight tenors, so we're looking for a premise that prevents tenors from being overweight. The second premise does it by making them effeminate - an effeminate tenor can't be overweight. The third premise gives us some effeminate tenors to work with. That means that we have some tenors who are effeminate and who are therefore not overweight. That makes this a perfectly valid argument.

• Your answer is quite helpful and explained ! Thanks a lot. Aug 23 '17 at 5:11
1. All tenors are either overweight or effeminate. No overweight tenor is effeminate. Some tenors are effeminate. Therefore some tenors are overweight.

The argument is invalid because existence in a category does not entail existence in its relative complement, and the first two statements tell us that is what are the categories.

$\begin{array}{ll:l}\because~&~T\subseteq O\cup E & \forall x~(Tx\to(Ox\vee Ex)) &\text{for tenors, effeminate and overweight are exhaustive categories} \\&~T\cap O\cap E=\varnothing & \neg\exists x~(Tx\wedge Ox\wedge Ex) & \text{for tenors, effeminate and overweight are exclusive categories} \\&~T\cap E\neq \varnothing~& \exists x~(Tx\wedge Ex) &\text{for tenors, the category of effeminate is not empty; there is some} \\\hline \not{\!\!\therefore\;}~&~ T\cap O\neq \varnothing & \exists x~(Tx\wedge Ox) \end{array}$

Consider a domain containing exactly one tenor, who is effeminate.   All three premises are satisfied therein, yet the conclusion is falsified.

1. All tenors are either overweight or effeminate. No overweight tenor is effeminate. Some tenors are effeminate. Therefore some tenors are not overweight.

The argument is valid because existence in a category exactly entails existence in the complement of its relative complement.

$\begin{array}{ll:l}\because~&~T\subseteq O\cup E & \forall x~(Tx\to(Ox\vee Ex)) &\text{for tenors, effeminate and overweight are exhaustive categories} \\&~T\cap O\cap E=\varnothing & \neg\exists x~(Tx\wedge Ox\wedge Ex) & \text{for tenors, effeminate and overweight are exclusive categories} \\&~T\cap E\neq \varnothing~& \exists x~(Tx\wedge Ex) &\text{for tenors, the category of effeminate is not empty; there is some} \\\hline \therefore~&~ T\smallsetminus O\neq \varnothing & \exists x~(Tx\wedge\neg Ox) \end{array}$

The first two claims of exhaustion and exclusion mean that a tenor is effeminate if and only if it is not overweight.   So if an effeminate tenor exists, an non-overweight tenor must (ie: the same tenor).