Using rules of inference, show If ∀x(Q(x)→(R(x)∧S(x))) and ∀x(P(x)∧Q(x)) are true, then ∀x(P(x)∧S(x)) is true

Question

Premise:

• ∀x(Q(x)→(R(x)∧S(x)))
• ∀x(P(x)∧Q(x))

Then show ∀x(P(x)∧S(x)) is true. I have been messing with this problem for well over an hour and I cannot find the rules of inference to prove this. I also cannot find too many examples online.

Any help is desperately needed.

Thanks

Answer 1

Hint

Use UI with 2nd premise to derive $Pa \land Qa$ and then use Simplification to derive $Pa$ and $Qa$.

Using again UI with 1st premise and then Modus Ponens, derive $Ra \land Sa$.

Use Simplification again and then Conjunction to derive $Pa \land Sa$.

The conclusion follows by Generalization.

Answer 2

I have been messing with this problem for well over an hour and I cannot find the rules of inference to prove this.

There are not that many rules of inference that might apply.

The premises have Universal Quantifiers, and Conjunction and Conditional connectives.

The rules of inference to use are clearly: Universal Elimination, Universal Introduction, Conjunction Elimination, Conjunction Introduction, and Conditional Elimination. Not in that order.

(You may know them by different names, but they should be recognisable. For example: you might know them as Universal Instantiation, Universal Generalisation, Simplification, Conjunction, and Modus Ponens, or something similar.)

Start from the conclusion and work back to the premises: You need to introduce a universal quantifier, so do that thing.$$\begin{array}{|l} \forall x~((Q(x)\to(R(x)\wedge S(x))\\\forall x~(P(x)\wedge Q(x)) \\\hline \quad\begin{array}{|l} [a]\\\hline \vdots\\ P(a) \wedge S(a)\end{array}\\ \forall x~(P(x)\wedge S(x))\hspace{10ex}\forall\mathrm I\end{array}$$ Now, how would you introduce that conjunction...?