I had a quiz today. Can someone check my answers? I got all false?
1. p -> (q->r) = p -> (q^r)
I got these are not equivalent
2. (p -> q) V (~p -> q) = q
I go this to be false, I actually got a tautology
Are these correct?
I had a quiz today. Can someone check my answers? I got all false?
1. p -> (q->r) = p -> (q^r)
I got these are not equivalent
2. (p -> q) V (~p -> q) = q
I go this to be false, I actually got a tautology
Are these correct?
1. correct
2. i would say true. the value of p doesn't matter, so the determining predicate is q.
I got true for the first one :S
I thought of it like this:
If a = b = c then a = both b and c since they are equivalent.. no?
if water is near then
if mosquitoes are around then
eggs are laid.
==
if water is near then
mosquitoes are around and eggs are laid.
I guess it could be false since the egg laying depends on whether or not mosquitoes are around. It's hard for me because it doesn't say IFF (If and Only If) mosquitoes are around.
God I hate this stuff..
For the second one, I thought of it like this:
If the gloves fit(p), the jury will acquit (q)) or (If they don't fit(p) the jury will acquit (q)) = the jury will acquit(q).
In either case, the jury acquits (q). So the statement is true. It doesn't matter if the gloves fit (p) or not, the jury acquits(q).
Last edited by Brandon; 02-14-2015 at 07:33 AM.
I am Ggzz..
Hackintosher
When in doubt, draw truth tables. That's like step one propositional logic debugging, right?
tl;dr: You got both answers correct (false, false), and correctly identified 2 as a tautology.Code:p q r (p -> (q -> r)) (p -> (q ^ r)) 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 1 p q ((p -> q) v (~p -> q)) 0 0 1 0 1 1 1 0 1 1 1 1
Last edited by core; 02-14-2015 at 08:38 AM.
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