What connective is missing in the following set of statements?

Question

Like here I have, say four statements which are related to same domain. They are P, Q, R and S and following set of logic is like this: (from some definition) P implies Q (from another definition) {P implies R which implies S} (These two statements in braces are interconnected) And then somehow Q implies S

My question is exactly how Q implied S. Aren't they both supposed to be equivalent like in = How Q is connected to S? what logic is applied here?

Here is the link of same question I have asked on meta. Reason for posting it on meta was that it's a crucial doubt and I'm missing something self evident logic here. What connective is missing in the following set of statements? As I have only 6 reps, I can't add pictures it's in meta section where I've added an image of it.

Answer 1

From premises:

$P \rightarrow Q$

$P \rightarrow R$

$R \rightarrow S$

it is indeed impossible to derive $Q \rightarrow S$

Part of the problem is that your $P$ and $Q$ are both referring to two atomic statements ... and you are 'losing' that fact by just using $P$ and $Q$

So, if you want to stick to propositional logic, I would symbolize something like $(a,b) \in R$ as an atomic statement in and of itself. That is, you could use:

$A$: $(a,b) \in R$

$B$: $(b,c) \in R$

$C$: $(b,a) \in R^{-1}$

$D$: $(c,b) \in R^{-1}$

$E$: $(a,c) \in R$

$F$: $(c,a) \in R^{-1}$

OK, so now you could represent your premises:

$A \land B$

$A \rightarrow C$

$B \rightarrow D$

$(A \land B) \rightarrow E$

$E \rightarrow F$

And the goal:

$(C \land D) \rightarrow F$

OK, but note that with this, you still cannot derive the conclusion.

And that is because the main problem with your proof is that as it is currently written, your written proof lacks the required flow. That is: you need to show that if $(b,a) \in R^{-1}$ and $(c,b) \in R^{-1}$, then $(c,a) \in R^{-1}$ (I assume you were asked to show that if $R$ is transitive, then $R^{-1}$ is transitive as well)

OK, so you should start with $(b,a) \in R^{-1}$ and $(c,b) \in R^{-1}$, then infer that $(a,b) \in R$ and $(b,c) \in R$, then use transitivity of $R$ to get $(a,c) \in R$, and from that you finally get $(c,a) \in R^{-1}$.

So, what do you need to do? Well, what you really have is that in general, $(x,y) \in R$ if and only if $(b,a) \in R^{-1}$. And so, you should really represent your givens as:

$A \land B$

$A \leftrightarrow C$

$B \leftrightarrow D$

$(A \land B) \rightarrow E$

$E \leftrightarrow F$

And now the goal:

$(C \land D) \rightarrow F$

is certainly derivable from those premises.

But I think the best way to do this proof logically, is by using first-order logic representations. As such, we are given that:

$\forall x \forall y (R(x,y) \leftrightarrow R^{-1}(y,x))$ (definition $R^{-1}$ as inverse of $R$)

and

$\forall x \forall y \forall z((R(x,y) \land R(y,z) \rightarrow R(x,z))$ ($R$ is transitive)

and the goal is:

$\forall x \forall y \forall z((R^{-1}(x,y) \land R^{-1}(y,z) \rightarrow R^{-1}(x,z))$ ($R^{-1}$ is transitive)

And here's a proof for that:

Answer 2

The logic applied from these kind of questions come from truth tables, and you use them to form a tautology.

Edit: Your link to meta looks like you're trying to form something called an equivalence relation, you're better off reframing your question and giving the equivalence relation and showing what steps you took to establish the reflexive, symmetric, and transitive properties.

Answer 3

My question is exactly how Q implied S.

Unless there's some additional logical statements that you didn't tell, it doesn't.

Counterexample:

• $P$ is the statement “$n=2$”.
• $Q$ is the statement “$n$ is an even number”.
• $R$ is the statement “$n$ is a prime number”.
• $S$ is the statement “$n$ is not a perfect square”.

Now from the definition of even numbers, you get $P\implies Q$ ($2$ is an even number).

From the definition of prime numbers you get $P\implies R\implies S$ ($2$ is a prime, and a prime is never a perfect square).

However $Q\implies S$ would be “An even number is not a perfect square“ which clearly is false, as proved by the counterexample $4$.