How to prove if two propositions are always true Let P1 and P2 denote the following  propositions: P1="CS is difficult or not many students like CS". P2="If math is easy, then CS is not difficult". Suppose that both P1 and P2 are true, determine if the following proposition is always true. Prove your answer. 
"Math is not easy if many students like CS".
For the answer, I wrote it in propositional terms :

P1 is : $(p \lor \lnot q)$ and P2 is : $(r \rightarrow \lnot p)$,

with $p$ denoting "CS is difficult", $q$ denoting "many students like CS", and $r$ denoting "math is easy".
Teacher told me that I wasn't supposed to use propositional logic or quantifiers, but did not elaborate. I know he wants some sort of mathematical proof, but he did not elaborate so I have no clue how to go about proving this.
 A: The following translation of (P1) might be a little problematic, but let's give this a try. We have a bunch of atomic statements:

(P) CS is difficult;
(Q) Many students like CS;
(R) Math is easy.

The exercise is to prove the following assertion:

Exercise. $\{(P \lor \lnot Q), (R \rightarrow \lnot P)\} ~\vdash~ (Q \rightarrow \lnot R)$.

Given that these P, Q, and R are propositional letters, and '$\lor$', '$\rightarrow$', and '$\lnot$' are propositional connectives, if a proof is required, then some logic of propositions must be used. Some options.

A. Formal derivation.

*

*$(P \lor \lnot Q)~~~~~~~~~~$ by hypothesis.


*$(R \rightarrow \lnot P)~~~~~~~~~$ by hypothesis.


*$Q \rightarrow P~~~~~~~~~~~~~~$ from (1) by the commutativity of $\lor$ & the meaning of $\rightarrow$.


*Suppose: $Q$.


*$|~P~~~~~~~~~~~~~~~~~~~~~~$ from (4) and (3) by modus ponens.


*$|~\lnot R~~~~~~~~~~~~~~~~~~~$ from (5) and (2) by modus tollens.


*$(Q \rightarrow \lnot R)~~~~~~~~~~~~$ from (4–6) by conditional proof.

There we're using some particular variant of the standard classical propositional calculus. We could give an informal version of the derivation as follows.

B. Informal proof.
Given (1) $(P \lor \lnot Q)$ and (2) $(R \rightarrow \lnot P)$, we want to show that $(Q \rightarrow \lnot R)$. To obtain the conclusion we assume $Q$, with the hope of proving $\lnot R$. Assume (3) $Q$. It will suffice to suppose $R$ and see if we can get a contradiction. Suppose (4) $R$. By premise (2), we know that $\lnot P$. But if $\lnot P$, then by premise (1), we know that $\lnot Q$. But $\lnot Q$ contradicts (3), so we must have $\lnot (4) \equiv \lnot R$. Since we had supposed $(3) \equiv Q$ and were able to obtain $\lnot R$, we can conclude that $(Q \rightarrow \lnot R)$ follows from (1) and (2), from $(P \lor \lnot Q)$ and $(R \rightarrow \lnot P)$.

While less formal, that one also depends on certain rules being present, like disjunctive syllogism and the methods of reductio and conditional proof. We can also proceed semantically:

C. Semantical proof.

Consider the 8-row truth-table T for propositional letters P, Q, and R, in that order. 6 rows of T (let's call it 'S' and store the selected rows in it: S := {1-4, 7-8}) satisfy $(P \lor \lnot Q)$, because only in rows 5 and 6 is $Q$ true but $P$ false. Among the six rows in S, only 4 rows satisfy $(R \rightarrow \lnot P)$, namely rows 2, 4, 7-8, so let's update S := {2, 4, 7, 8}. To prove the consequence we have to check if all the rows in S satisfy $(Q \rightarrow \lnot R)$, which is false only when $Q \equiv R \equiv \top$. But $Q$ and $R$ have different truth-values for the rows in S, so all the rows in S satisfy $(Q \rightarrow \lnot R)$. This means that any assignment of truth-values to P, Q, R that satisfies the two premises, will also satisfy the conclusion.
There are, of course, other ways to go, but one of these might satisfy your teacher.
