# Using the rules of inferences

I know the rules of inferences and logical equivalence but I cannot seem to validate this argument. I rewrote the first premise as $\neg p\vee q$ other from that I am stuck. Any help will be appreciated.

$$\begin{array}{c} p \to q \\ (q \land r) \to s \\ r \\ p \\ \hline s \end{array}$$

## 2 Answers

Since $p$ and $p\to q$, we have $q$ by modus ponens.
Since $q$ and $r$, we have $q \land r$ by definition of conjunction.
Since $q \land r$ and $(q \land r) \to s$, we have $s$ by modus ponens.

1. Assume: $((p \implies q) \land ((q \land r) \implies s) \land r \land p)$

2. $(p \implies q)$ ($\land E$ 1)

3. $((q \land r) \implies s)$ ($\land E$ 1)

4. $r$ ($\land E$ 1)

5. $p$ ($\land E$ 1)

6. $q$ ($\mathord{\Longrightarrow} E$ 5,2)

7. $q \land r$ ($\land I$ 4,6)

8. $s$ ($\mathord{\Longrightarrow} E$ 7,3)

Therefore: $((p \implies q) \land ((q \land r) \implies s) \land r \land p) \implies s$

Notations: $\land I$ is the introduction rule of conjunction, $\land E$ is the elimination rule of conjunction. $\mathord{\Longrightarrow}E$ is the elimination rule of implication, i.e. modus ponens.