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Anonymous Mon 13 Sep 15:24:56 2021 No.13636306 View Reply Original Report

Quoted By: >>13636312 >>13638749

I'm a huge retard and can't seem to use basic tricks to find logical equivalents. Could someone help me through this one, with some description. I find that the rules that I can use are not sufficient, if I don't know algebra well (which I don't)

p (q r) ? ( q /\ p ) r

Sincerely, OP.

Anonymous

Anonymous Mon 13 Sep 2021 15:25:53 No.13636312 Report

Quoted By:

>>13636306
p -> (q -> r) ? ( q /\ p ) -> r

Guess it couldn't use the operators. So made them like this instead

Anonymous

Anonymous Mon 13 Sep 2021 15:38:05 No.13636349 Report

Quoted By: >>13636361

p->(q->r)
= p -> (~q V r)
= ~p V (~q V r)
= ~p V ~q V r
=~(p /\ q) V r
= (p /\ q) -> r
= (q /\ p) -> r

remember to get vaccinated

Anonymous

Anonymous Mon 13 Sep 2021 15:41:54 No.13636361 Report

Quoted By: >>13636374

>>13636349
So you use conditional twice in a row.
Then you use distributive to get the "not" behind
and then you use reverse conditional
and then commutative to reverse p and q.

Have I understood this correctly?

Cheers, that was stupidly easy. I'm a dumb dumb.

Anonymous

Anonymous Mon 13 Sep 2021 15:45:54 No.13636374 Report

Quoted By: >>13636382

>>13636361
I use the identity that p->q = ~p V q

What you are referring to as "distributive" to get the "not" behind is known as de Morgan's law.

Namely: ~p V ~q = ~(p /\ q) . I understand you are trying to associate this to regular algebra by calling it distributive property, but I don't want to mislead you. This is de Morgan's law.

once you've applied de Morgan's law, you are able to use the same p->q = ~pVq identity to get the answer you are looking for.

Anonymous

Anonymous Mon 13 Sep 2021 15:50:23 No.13636382 Report

Quoted By:

>>13636374
Yeah I see now, it's de Morgan's, not distributive. Of course...

Thanks a lot!

Anonymous

Anonymous Tue 14 Sep 2021 05:43:06 No.13638749 Report

Quoted By:

>>13636306
In lambda-calcululs ("\" stands for lambda), construct a proof of (P -> Q -> R) -> ((Q * P) -> R) as \f -> \(a,b) -> f a b and construct a proof of ((Q * P) -> R) -> (P -> Q -> R) as \f -> \a -> \b -> f (a,b). This maps to a natural deduction style proof rather simply via the Curry-Howard correspondence.

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