1. ## logic question

I have to show "a" occurs if & only if, whenever "b" occurs, "c" occurs.

For the only if direction I'm pretty sure this means if I have to assume "a" & "b" and then use this to prove "c".

But for the if direction, is it correct that I have to assume "c" & "b" and use this to prove "a"?

2. the if part can be written (b -> c) -> a.
Since you can rewrite (b -> c) as (!b | c)
to prove the if part you have to prove
!b -> a
c -> a

3. Thanks, can you define your "!" and "|" symbols for me?

4. Originally Posted by RanDom
I have to show "a" occurs if & only if, whenever "b" occurs, "c" occurs.

For the only if direction I'm pretty sure this means if I have to assume "a" & "b" and then use this to prove "c".

But for the if direction, is it correct that I have to assume "c" & "b" and use this to prove "a"?

$(b\,\Rightarrow\,c)$ is equivalent to $(\neg\,b\ \mbox{or}\ c).$ So to prove $(b\,\Rightarrow\,c)\,\implies\,a,$ you have to show that both $\neg\,b\,\implies\,a$ and $c\,\implies\,a.$

5. i used a notation similar to the one of the java programming language
& = logical conjunction (∧)
! = logical negation (¬)
| = logical disjunction (∨)

6. A little off topic, - how would you show that, (for the only if direction), a -> (b ->c) is equivalent to (a & b) -> c ?

And just to clarify,
Originally Posted by RanDom
for the if direction ... I have to assume "c" & "b" and use this to prove "a"
is incorrect, right?

7. Originally Posted by TheAbstractionist
$(b\,\Rightarrow\,c)$ is equivalent to $(\neg\,b\ \mbox{or}\ c).$ So to prove $(b\,\Rightarrow\,c)\,\implies\,a,$ you have to show that both $\neg\,b\,\implies\,a$ and $c\,\implies\,a.$
Wait... I know for a fact that c implies a is false... !!!

8. always consider x -> y is equivalent to !x | y

so (a & b) -> c
is equivalent to !(a & b) | c
and this is equivalent to !a | !b | c

also a -> (b ->c)
is equivalent to !a | (!b | c)

9. Originally Posted by RanDom
Wait... I know for a fact that c implies a is false... !!!
It can’t be false if $(b\,\Rightarrow\,c)\,\implies\,a$ is to hold.

The implication is equivalent to $\neg\,(b\,\Rightarrow\,c)$ or $a.$

$\neg\,(b\,\Rightarrow\,c)$ is equivalent to $\neg\,(\neg\,b\ \mbox{or}\ c)$ which is equivalent to $b\ \mbox{and}\ \neg\,c.$

Hence the implication is equivalent to $(b\ \mbox{and}\ \neg\,c)$ or $a.$ It follows that if $c$ is true, then $(b\ \mbox{and}\ \neg\,c)$ is not true and so $a$ must be true; in other words, $c\,\implies\,a.$

10. Yep nevermind... there was a subtle difference for c->a and the result I knew was false... Thanks for your help guys

11. Hello, RanDom!

Have interpreted the question correctly?

I have to show "a" occurs if & only if, whenever "b" occurs, "c" occurs.
I could be wrong . . .

It says: . $a \:\longleftrightarrow \b \to c)" alt="a \:\longleftrightarrow \b \to c)" />

So we must prove: . $\begin{array}{c}a \longrightarrow (b \to c) \\ \text{and} \\ (b \to c) \longrightarrow a \end{array}$