in Mathematical Logic edited by
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Use rules of Inference to show that the hypotheses

"If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on."

"If the sailing race is held, then the trophy will be awarded." and

"The trophy was not awarded" imply the conclusion

"It rained."

I worked it like below

p : It rains

q: It is foggy

r: Sailing race will be held

s: Life saving demonstration will go on

t: Trophy will be awarded

(1) $(\sim p \lor \sim q) \rightarrow (r \land s)$

(2)$r \rightarrow t$

(3)$\sim t$

From (2) and (3) I get $\sim r$ means $r$ is false.

(1) can be re-written as

$((\sim p) \rightarrow (r \land s)) \land ((\sim q)\rightarrow(r \land s))$....................(4)

Since r is false and (1) if it is true, then the LHS of implication has to be false.

So, it means it comes as $\sim(\sim p \lor \sim q) \equiv p \land q \,will\,be\,true$

WHich means it both rained and it was foggy. Can we have 2 conclusions in this?
in Mathematical Logic edited by
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there is no restriction that every time you get that only one conclusion.

in a running race, A , B and C are participating, winner given that A is winner.... isn't this concludes remaining all runners ( B , C ) are loosers.


 

(1) (∼p∨∼q)→(r∧s)  ≡ ~ ( p ∧ q ) → ( r ∧ s )

(2)r→t

(3)∼t

 

from (3) and (2) concluded that ~r -----(4)

From (4) you can conclude that  ( r ∧ s ) is False in ( 1 )  ===>  ~ ( p ∧ q )  is False  ===> ( p ∧ q ) is true

∴ p is true and q is true

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1 Answer

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Yes it is correct. You can have more than one conclusion. You got p AND q. By rule of simplification bot p and q is true. Using rules of inference we can filter out the invalid rules and it doesn't mean your conclusion must be unique.

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