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Using only the rules of inference and the logical equivalences, show that the following argument is valid. You may assume that all the premises given are true.

Premises:

𝑢 ∧ 𝑡

𝑟 → 𝑞

s ∨ (𝑝 → 𝑟)

¬𝑠 ∧ 𝑡

¬𝑞 ∧ 𝑢

Show: ¬p

So I've attempted the solution but I'm stuck and not sure what steps I should take next. Any input or help would be appreciated.

My solution so far:

1) 𝑢 ∧ 𝑡

2) 𝑟 → 𝑞

3) s ∨ (𝑝 → 𝑟)

4) ¬𝑠 ∧ 𝑡

5) ¬𝑞 ∧ 𝑢

6) ¬𝑠 by simplification (4)

7) (𝑝 → 𝑟) by disjunctive syllogism (3, 6)

8) (𝑝 → q) by hypothetical syllogism (2, 7)

9) ¬𝑞 by simplification (5)

10) ¬p by modus tollens (8, 9)

I'm trying to get rid of 1) and 6) left but I'm out of formulas I can think of to use. I thought maybe I could use addition somehow, but I'm not quite sure how that formula works. Like would that allow me to add any variable from the alphabet that I want or can I only use the variables that have already been defined?

Any thoughts or ideas on what steps I could take next would be appreciated.

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  • $\begingroup$ Not clear... It works: you have derived $\lnot p$. $\endgroup$ – Mauro ALLEGRANZA May 28 at 7:13
  • $\begingroup$ don't I have to get rid of the other premises? I thought I could only have 1 statement left. $\endgroup$ – GilmoreGirling May 28 at 7:15
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    $\begingroup$ Not necessarily; it means only that premise 1) is not needed for the argument. $\endgroup$ – Mauro ALLEGRANZA May 28 at 7:31
  • $\begingroup$ 1 is indeed redundant. It follows from 4 and 5. $\endgroup$ – PJK May 28 at 7:33
  • $\begingroup$ @PJK A redundant premise is merely one that is not required to entail the conclusion. $\endgroup$ – Graham Kemp May 30 at 2:20
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$$\dfrac{\dfrac{\lower{1.5ex}{r\to q}~~\dfrac{\lower{1.5ex}{s\lor (p\to r)}~~\dfrac{\lnot s\land t}{\lnot s}{\tiny\text{simplification}}}{p\to r}{\tiny\text{disjunctive syl.}}}{p\to q}{\tiny\text{hypothetical syl.}}~~\dfrac{\lnot q\land u}{\lnot q}{\tiny\text{simplificaton}}}{\lnot p}{\tiny\text{modus tolens}}$$


You have correctly proven that $r \to q, s\lor (p\to r),\lnot s\land t,\lnot q\land u\vdash \lnot p$ and that was all you needed to do.

As the commenters have stated, the fact that you do not need premise $u\land t$ to derive $\lnot p$ is not a problem.   It is just a redundant premise.   You may have as many redundant premises as you wish.

$$m\lor n~,~u\land t~,~ r \to q~, s\lor (p\to r)~,\lnot s\land t~,\lnot q\land u\vdash \lnot p$$

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