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another way to express that <=> part is to split it up both ways: -q -> -p -p -> -q
-q <-> -p is only true if bith those expressions are true
I meant to say *and p is F. Sorry. So will the value be false?
dunno, id have to remember all the assorted details :)
don't you love truth tables?
p -p q -q -p -> -q -q -> -p -q <-> -p 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 1 1 aint that the truth :)
if p is 0 then -p is 1
other than that, im sure I cant undersatnd the rest of the question
we also have \[\lnot p \leftrightarrow \lnot q \equiv p \leftrightarrow q\]
which certainly means if \[\lnot p \leftrightarrow \lnot q \] is true then \[q\rightarrow p\] is true
If ~q <=> ~p is T p -p q -q -p -> -q -q -> -p -q <-> -p 1 0 1 0 1 1 1 0 1 0 1 1 1 1 and p is F, p -p q -q -p -> -q -q -> -p -q <-> -p 0 1 0 1 1 1 1 find the truth value of q =>p. p q -q <-> -p 0 0 1 0 -> 0 = 1
wow! but i still think the question is odd. if p iff q, then q implies p and p implies q, really has nothing to do with a particular case
indeed, its just a practice drill :)
Wow. Thanks for breaking this down. My prof doesn't go this in-depth. Guess he gets bored since he already knows this stuff haha
im in calc 3 and my teacher is still teaching us ow to add and subtract fractions; since there are students in class that have no idea how to .... makes me wonder how they even got there to begin with
then she wonders why she gets behind in the lessons plans :)
do you have to put the fractions over the same denominator? or can you just add straight across?
i just add straight across and hope for the best lol
Remarkable. Talk about breaking away from the syllabus..sheesh