The discussion centers on the implications of the formula $$p\wedge\neg (q\vee r)$$ in propositional calculus. Participants analyze whether this formula implies either $$(p\wedge\neg q)\vee (p\wedge\neg r)$$ or $$(p\wedge\neg q)\wedge (p\wedge\neg r)$$. The consensus is that the formula does imply the first option, as it logically follows from the negation of the disjunction. The second option does not hold as a valid implication.
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Monoxdifly said:$$p\wedge\neg (q\vee r)$$
=$$p\wedge(\neg q\wedge\neg r)$$
=$$(p\wedge\neg q)\wedge (p\wedge\neg r)$$
So, it implies 2)
solakis said:Sorry,but i am trying to find out how you got $$(p\wedge\neg q)\wedge (p\wedge\neg r)$$ from
$$p\wedge(\neg q\wedge\neg r)$$
Olinguito said:And ($\wedge$) is distributive over itself since it is associative, commutative and idempotent: https://proofwiki.org/wiki/Associative_Commutative_Idempotent_Operation_is_Distributive_over_Itself.
Aren’t the expressions in red equal?solakis said:(a*b)*(a*c)= a*[b*(a*c)]
Now where wiki gets : (a*b)*(a*c)=a*(b*a)*c ??
Olinguito said:Aren’t the expressions in red equal?
solakis said:By using associativity you can get them equal.
But i am asking ,how wiki gets, a*(b*a)*c from (a*b)*(a*c) by ‹using associativity
Parenthesis are very important in this kind of proof otherwise a lot of wrong proofs can be given
Look at the following proof:
(a*b)*(a*c)= (a*a)*b*c=............By commutativity
=a*b*c.................By indepondency
I like Serena said:If we do not have associativity, parentheses are indeed very important.
If we do have associativity, parentheses are redundant, and often left out.
In this case we have associativity, don't we?
solakis said:Can you give an example of a system with a binary opperation where by using associativity you can prove that parenthesis are redundant
Monoxdifly said:$$p\wedge\neg (q\vee r)$$
=$$p\wedge(\neg q\wedge\neg r)$$
=$$(p\wedge\neg q)\wedge (p\wedge\neg r)$$
So, it implies 2)