Prove following argument is valid

Question

Prove the following argument is valid.

If Ralph doesn't do his homework or he doesn't feel sick, then he will go to the party and he will stay up late. If he goes to the party, he will eat too much. He didn't eat too much. So Ralph did his homework.

So far I came up with the following:

Let's Set:
H(x): Ralph does his homework
S(x): Ralph feels sick
P(x): Ralph goes to the party
L(x): Ralph stays up late
E(x): Ralph eats too much

Argument:

(¬H(x) v ¬S(x)) → (P(x) ∧ L(x))
P(x) → E(x)
¬E(x)
H(x)

Not positive the above is right or not. But either way - I don't understand how to do the proofs. We can use the following Rules of Inference:

Modus Ponens
Modus Tollens
Hypothetical Syllogism
Addition
Simplification
Conjunction
Disjunctive Syllogism

Help would be much appreciated.

Answer 1

I hope that my mobile phone manages what I want to share. You correctly transformed the argument in the language of logic but I will change something:

$1. (\lnot H \lor \lnot S) \to (P \wedge L) \\ 2. P \to E\\ 3. \lnot E\\ 4. \therefore H$

The entries 1 to 3 are the premises which you are allowed to use to prove the conclusion 4. I use only big capitals because I am writing on a mobile device and you do not need those "(x)" because you have no predicates with more than one entry, propositional logic works fine enough.

Then you can use the rules to prove your statement 4:

$$5. \lnot P \text { (modus tollens: 3,2)} \\ 6. \lnot P \lor \lnot L \text { (addition: 5)} \\ 7. \lnot (P \wedge L) \text { (DeMorgan: 6)}\\ 8. \lnot (\lnot H \lor \lnot S) \text { (modus tollens: 7,1)}\\ 9. H \wedge S \text { (DeMorgan: 8)}\\ 10. H \text { (simplification: 9)} $$

Use the rules to transform your premises until you have the desired conclusion.

Answer 2

By Direct Proof :

$1. (\sim H(x) \lor \sim S(x)) \rightarrow (P(x)\lor L(x))$

$2. P(x) \rightarrow E(x)$

$3. \sim E(x)$
$---------------------$

$4. H(x)$

$5. P(x) \rightarrow E(x) \equiv \quad\sim P(x) \lor E(x)$ by Material Implication

$6. \sim P(x)$ , #5 and #3 by Disjunctive Syllogism

$7. \sim P(x) \lor \sim L(x)$ , #6 by Addition ( I just add $ \sim $L(x))

Since #7 is logically equivalent to $\sim(P(x) \lor L(x))$ by De Morgan's Law,

$8. \sim (\sim H(x) \lor \sim S(x))$ , #1 and #7 by Modus Tollens.

Distributing the $\sim$, we'll have,

$9. H(x) \land S(x)$ by De Morgan's and Double Negation

$10. H(x)$ by Simplification $\qquad \blacksquare$