# Show if this is valid or invalid (Propositional Logic)

I just don't understand clearly what the question wants me to do?

It says: Check if the following is valid or invalid (I will take one question as an example.) $$H\implies D$$ $$R\implies S$$ $$Therefore: (H \land R \implies S \land D)$$ I understand that they want me to show a proof of this conclusion and also to show if this is valid. Just looking at it, it seems valid, but how do I prove it using natural laws of deduction (the logical way)? And just for information, I know all the laws of deduction. I just don't know how to apply them in situations like this.

• Do you know how to make truth tables? One way you can do is it by making the truth table for $((H \implies D) \land (R \implies S)) \implies ((H \land R) \implies (S \land D))$. – Lee Mosher Sep 24 '17 at 16:39
• "laws of deduction" ? What ones ? Natural Deduction ? Logical equiv ? – Mauro ALLEGRANZA Sep 24 '17 at 16:49
• With Nat Ded is pretty simple: Assume $H \land R$ and derive $D$ and $S$ separately; then use $\land$-intro. – Mauro ALLEGRANZA Sep 24 '17 at 16:50

There are many 'laws of deduction': there are many different systems of deduction, each of which with their own set of laws or rules ... so it would be good to know which rules you are allowed to use.

Nevertheless, here is a proof using fairly commonly used rules:

Bram28's natural deduction proof shows that it is valid. However, suppose it were not valid. One way to check if it were invalid is to try to find a counter example that would make all of the premises true but the conclusion false.

To try to find such a counter example assign truth values so that the conclusion is false. A false conclusion would require that both $$H$$ and $$R$$ be true and at least one of $$D$$ and $$S$$ be false. With those values can one make all of the premises true? If both $$H$$ and $$R$$ are true then at least one of the two premises would be false, the one where either $$D$$ or $$S$$ was false. Hence there is no valuation of $$H$$, $$R$$, $$D$$ and $$S$$ that would make this invalid.

Alternatively, one could put this in a truth table generator. To do that conjoin the premises to form the sentence $$(H\to D) \land (R\to S)$$. Connect that sentence to the conclusion $$(H \land R)\to (D \land S)$$ using a conditional ($$\to$$) like so: $$((H\to D) \land (R\to S)) \to ((H \land R)\to (D \land S))$$

Any column reading false below that conditional connective would have valuations that make the conditional false. As it turns out, and as we can expect from the already existing proof, there aren't any:

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html