I don't think we have "assumption" we have laws like domination law, identity law, double negation law, De Morgans Law, etc

I had the same book, I'm pretty sure. What page/problem is that?

I recall you can turn p -> r into something else like (not p v r) or something. Do that with both implications on the left...might get you started.

I probably did that problem. I'll have to check.

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its on a worksheet for class. I just dont understand how you can change one thing into another

I once found a quarter in "Discrete" and almost got ran over by some old lady.

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Check out page 25 in the book, it'll give you a table of logical equivalences.

Here is the solution to your posted problem:


What you have to do, and this applies to most of the problems, is rewrite the implication as an AND/OR logical equivalence (page 25) then, regroup items, add items in, and manipulate them to get them in the form of the RHS.

You can re-write things because they have the same truth table values as other statements.

One of the important rules in discrete is DeMorgans laws. Read the write up on page 25 in the book, they lay it out nicely and have examples that will be helpful.

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When I went to school we had to first prove DeMorgan's law in order to use it. In any case, first step remains the same. Prove it going one way, then prove it going the other. You just have to figure out the equivalents just like in algebra.

Here's a tip. Anything inside parantheses, for example (p -> r) v (q -> r) you can try to look at it as if everything inside the parantheses is just one. So think:

(p -> r) becomes the P in DeMorgan's law and (q -> r) becomes the Q in Demorgan's law.

Demorgan's law is -P or -Q is equivalent to -(P and Q)

Since P is equivalent to --P (double negation) or if you want to look at it like -notP

-notP or -notQ is equivalent to -(notP and notQ)

Hence

(p -> r) v (q -> r) by DeMorgan's law becomes -(-(p -> r) AND -(q -> r))

You really only need three rules plus negation for any proof. Material implication which changes a conditional to a disjunction and both DeMorgan's laws. That is the sum total of Boolean Logic.

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its helping.. a little.. but I just cant seem to still get it.

Prove (p → r) ∨ (q → r) ≡ (p ∧ q) → r

but since there are no ~ De Morgans law wont come into play here.. I think.

So I have to make the right equal to the left. so I can leave the left alone, correct?

You got to learn treat what's inside each well formed parenthesis as a single unit. The opposite of what is done in arithmetic... sorta. Don't take that as a rule

Edit: You need to start by understanding DeMorgan's.

The negations are screwing with your head. Do this exercise and use DeMorgan's on the following. Note the double and triple negations.

--P or --Q is equivalent to?

--(P and Q) is equivalent to?

-(--P and -Q) is equivalent to?

And so on.

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DeMorgans comes into play later down in the problem.

You have to re-write p->r and q->r into basic and/or form. See my post a few back.

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so step 2.. is that considered "definition of a conditional"?

Actually yes. Whenever you see a conditional rewrite it using Material Implicaition theorem.

P -> Q is equivalent to -P or Q

The same thing applies with the negations.

So -P -> Q is equivalent to --P or Q. Remove the negations and --P or Q is equivalent to P or Q.

Similiarly P -> -Q is equivalent to -P or -Q.

Once you convert it to Boolean form you can use DeMorgan's theorem to fiddle around with it and then reconvert to conditional form. The way you do it is exactly like trigonometric identities.

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so would it go

definition of conditional

distributive law

demorgans law

then we have our equivelence?

Is that the name of the rule in the book?

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well there is an example (#3 pg 23) showing p->q and ~p v q are logically equivalent

If I were to look James' solution it only shows half the solution. You have to prove it the other way.

P -> Q AND Q -> P becomes P is equivalent to Q.

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