Anyone ever take a discrete math course? I'm stumped beyond infinity.

is it linear? groups?

effin' hate that 5hit!

hey... wait a second... you can't have infinity in discrete math...

Nope. Soon as the numbers go above ten my brain shuts down.

- Peter.

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Peter, take your shoes off & you will be good for 20 !!!
Don, infinity is just an 8 that fell over !!

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Yes, I took one. It was alright, it was mainly logic, and proofs of various items. Only math course I've ever gotten anything other than an A in..

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"The law, in its majestic equality, forbids the rich as well as the poor to sleep under bridges, to beg in the streets, and to steal bread."

its killing me. I get the idea of whats going on but dont understand how to do the proofs

Which book are you using for logic?

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Discrete Mathmatics and Its Applications sixth edition Kenneth H. Rosen

Hmm... never used the system. What kind of rules are you allowed so far, are you using quantifier logic, and are you allowed to use reductio ad absurdium?

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You don't need a weatherman to know which way the wind blows - Robert A. Zimmerman

quantifiers, laws of disjuntion/conjunction

here is an in class example

Prove (p → r) ∨ (q → r) ≡ (p ∧ q) → r using logical equivalence. Please the name
of the laws you use.

I've got no idea how to start.

Don, I actually recognize that from over 35 years ago!! There are a series of laws, if you start with the RHS you can use the laws to change it a couple of times, you will then be able to make it look like the LHS & so proved

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I no longer question authority, I annoy authority. More effect, less effort....

1967 230-6 auto parts car. rust bucket.
1980 300D now parts car 800k miles
1984 300D 500k miles
1987 250td 160k miles English import
2001 jeep turbo diesel 130k miles
1998 jeep tdi ~ followed me home. Needs a turbo.
1968 Ford F750 truck. 6-354 diesel conversion.
Other toys ~J.D.,Cat & GM ~ mainly earth moving

yeah.. I dont follow how to actually do that. it doesnt click

so I ignore the left hand side all together and make the right look like the left.. somehow?

I'll get you started.

Since you are proving equivalence you need to prove on direction then prove the other.

First prove [(p → r) ∨ (q → r)] -> [(p ∧ q) → r ]

Assume one side of the conditional

1) (p → r) ∨ (q → r) (assumption)

Now you need to prove the other side which is a conditional also. So you assume the left side of the conditional as well. Indent one time. I can't seem to do it on the forum.

2) (p ∧ q) (assumption)

Now you need to prove r.

3) -r (assumption)

.
.
.

Once you've prove the above conditional you need to prove the conditional the the other way. (since p ≡ q is p -> q and q -> p right?)

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