The discussion revolves around the implications of a formula in propositional calculus, specifically examining whether the expression $$p\wedge\neg (q\vee r)$$ implies either $$(p\wedge\neg q)\vee (p\wedge\neg r)$$ or $$(p\wedge\neg q)\wedge (p\wedge\neg r)$$. The focus is on logical reasoning and proof techniques within the context of propositional logic.
The discussion is currently unresolved, with no consensus reached on the implications of the formula.
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)