help verifying my answer for this" premise-conclusion" question

Question

For each of the premise-conclusion pairs below, give a valid step-by-step argument (proof) along with the name of the inference rule used in each step.

(a) Premise: {¬p ∨ q → r, s ∨ ¬q, ¬t, p → t, ¬p ∧ r → ¬s}, conclusion: ¬q

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I did this:

1) $ p → t $ Premise

2) $¬t $ Premise

3) $¬p $ Modus Tol.

4)$¬p ∨ q → r$ premise

5) $q → r $ disj. syll. from 3 , 4

6) $q ∨ ¬ r $ implication law ( don't know if I'm allowed to use that here)

7) $¬ ( q ∨ ¬ r ) $ Don't know the name of this rule. I'm thinking negation?

8) $¬ q ∧ r $

9)$ ¬ q $ by simplification.

Answer 1

You misused the ~p to derive "q->r" given that you have ~p and can only derive ~p from that specific disjunction. I'm not sure what you did in step six either.

Perhaps try:

Premise: p -> t Premise: ~t

[~p v t], DeMorgan's law, therefore ~p given premise ~t and disjunctive syllogism.

Premise: [~p v (q -> r)],

use negation [~(~p v (q -> r))] and distribute negation for:

[p & ~(q -> r)] and distribute after converting conditional to disjunction to derive:

[~(~q v r)] then [q & ~r].

Thus we have [p, ~p, q, and ~r].

with p proven, we can revisit your fourth premise, derive the conditional and use the ~r derived from my proof with modus tollens to prove ~q.

Answer 2

$\neg p \vee q \to r\; $ is $\; \big((\neg p) \vee q\big) \to r \;$ and not $(\neg p) \vee (q\to r)\;$.

Operator precedence goes to the disjunction before the implication.

$\begin{array}{lll} 1) & p \to t & \text{ Premise } \\ 2) & ¬t & \text{ Premise } \\ 3) & ¬p & \text{ Modus Tolens} \\ 4) & ¬p∨q→r & \text{Premise} \\ 5') & \neg p \vee q & 3, \text{disjunction introduction} \\ 6') & r & 4, 5', \text{modus ponens} \\ \vdots & \vdots & \vdots \end{array}$

Can you finish from this?

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PS: You also have the implication equivalence incorrect in your original 6; it's the negation of the antecedent disjunct with the consequent, not the antecedent disjunct with the negation of the consequent.

$$\; a\to c \iff \neg a \vee c \;$$

PPS: You don't need implication equivalence to solve this question. But learn what it is for when you do.