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Let

P – Light is on

Q – The switch is down

R – The door is open

If the switch is down then the light is on.
If the switch is not down then the door is open.
If the door is open then the light is on.

Therefore the light is on;

Prove or disprove the argument using, i. Rule of inference ii. Truth table

\ My answer .....

\ Rule of inference

A. Q⇒P
B. ~Q⇒R
C. R⇒P

D. ~Q⇒P [By {B} and {C}]
E. P [By {A} and {D}]

Truth Table

Which is different from the answer I get from the rule of inference. Can anyone tell me where did I go wrong? Thanks!!

Glorfindel
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Padmal
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2 Answers2

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Your truth table supports the result of the proof: indeed, in all rows where all the premises are true, the conclusion is also true.

Perhaps you're mixing the situation up with one where you try to prove that $P$ is a tautology?

  • So the answer is "P is the correct conclution" right? :) Even in the truth table, if I get the product of 1 and 4 other than getting the product of 1, 2 and 3; Result is the same as the "P" colomn. So can we conclude that "P" is a valid conclusion from the premises given? – Padmal May 29 '15 at 04:30
  • @Blogger The last column does not have to be exactly the same as $P$, you only need $1$ in the first column whenever there is a $1$ in the last column. You are only proving a claim about the state of $P$ whenever the three premises all hold; you don't care at all what its state is otherwise. – Graham Kemp May 29 '15 at 04:40
  • So it is like; IF " Q→P (and) ¬Q→R (and) R→P = 1" THEN "P = 1"? – Padmal May 29 '15 at 04:43
  • @Blogger It's exactly like that. – Graham Kemp May 29 '15 at 04:46
  • Hey thanks alot @GrahamKemp Now I get it. I thought both the rows should be always identical. Thanks again you saved my day! :D – Padmal May 29 '15 at 04:48
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No, your table is correct.   You may be interpreting the result wrong.   You wish to have $P$ true whenever the statements $Q\to P, \neg Q\to R,$ and $R\to P$ are all true at the same time.   That happens on the last three rows, and $P$ is true for each one. $$Q\to P, \neg Q\to R, R\to P \;\vdash\; P$$

PS: Your application of rules of inference is okay too.   You used hyperthetical syllogism and disjunctive elimination.

Graham Kemp
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  • Thanks for the reply. Do you mean that the whole "P" row doesn't need to be identical to the row we get after Q→P (and) ¬Q→R (and) R→P ?? Because in the truth table there is a 0 in the 5th line where it is 1 in the "P" row :( – Padmal May 29 '15 at 04:35
  • @Blogger That's it. The last column does not have to be exactly the same as $P$, you only need $1$ in the first column whenever there is a $1$ in the last column. You only wish to prove a claim about the state of $P$ whenever the three premises all hold; you don't care at all what its state is otherwise. – Graham Kemp May 29 '15 at 04:43