Show $\neg R \vee C, (C \wedge R) \to \neg R, (\neg(R \vee R) \to R)$ are jointly consistent.

Question

Working on P.D. Magnus. forallX: an Introduction to Formal Logic (pp. 183, exercise B. 10).

I would need to find a single row in its truth table where this sentence is true:

$$(\neg R \vee C) \wedge ((C \wedge R) \to \neg R) \wedge (\neg(R \vee R) \to R)$$

However, I've not succeedded in finding one. Perhaps, am I missing something ?

Answer 1

That's impossible. For the third to be true, $R$ must be true. So for the first to be true as well, $C$ must be true too. But that valuation makes the second false.