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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.

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

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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.

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  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.

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