I often have trouble understanding why A $\displaystyle \Rightarrow$ B means A only if B and also why the contapositive is equivalent. Can anyone give me some tips or useful ways of thinking about it? Thanks
$\displaystyle A \Rightarrow B$ means "if A is true then B is true." contrapositive is "if B is not true then A is not true."
to see why the contrapositive is equivalent assume the contrapositive statement holds. that is "if B is not true then A is not true.". we shall show that this implies "if A is true then B is true." use contradiction method. so further assume that there exists a case when A is true but B is not true. BUT WE STARTED WITH if B is not true then A is not true(this was the contrapositive statement ). so there is a contradiction between the blue and the green statements.
hence "if B is not true then A is not true" leads us to "if A is true then B is true."
now it remains to show that "if A is true then B is true" leads us to "if B is not true then A is not true."
if this is done then the equivalence of the contapositive and the original will be established. can you try and do it?
assume A does not imply B. So A can be true and B can be false which contradicts
not B $\displaystyle \Rightarrow $ not A. Conclude A $\displaystyle \Rightarrow$ B
Now let A $\displaystyle \Rightarrow$ B. Assume 'not B' does not imply 'not A' so B can be false and A can be true. Again there is a contraction. Conclude 'not B' implies 'not A'
I m very reluctance to get join the discussion.
But first, why not learn to post in symbols? You can use LaTeX tags
[tex]A \Rightarrow B[/tex] gives $\displaystyle A \Rightarrow B $
Second, I think you are being ‘loose’ in what implies means.
This is the case: $\displaystyle \left( {A \Rightarrow B} \right) \equiv \left( {\neg A \vee B} \right)$.
Now when you say “assume A does not imply B” you are saying “it is false that A does imply B”.
That is, $\displaystyle \neg \left( {\neg A \vee B} \right) \equiv \left( {A \wedge \neg B} \right)$.
/\ means conjunction ("and"); LaTeX: \land or \wedge
\/ means disjunction ("or"); LaTeX: \lor or\vee
≡ means biconditional ("if and only if"), also denoted by <=> and <->; LateX: \equiv, \Leftrightarrow, \leftrightarrow
¬ means negation; LaTeX: \neg
For simple situations, I prefer to use ASCII characters /\, \/, ->, =>, ~ (for negation).
Edit: There is also a list of logic symbols in Wikipedia, from where Unicode symbols can be copied.
The code [TEX]A\not\Rightarrow B[/TEX] yields
$\displaystyle A\not\Rightarrow B.$
The \not command, in general, puts a slanted slash through the subsequent operator.
Could be wrong, but I don't think there is any difference between 'A implies B is false' and 'A does not imply B'.Secondly Plato said that I was really saying that 'A implies B is false'. What is the difference between this and 'A does not imply B'.
Since I don't recall seeing anyone respond to that part of your question, I'll sign in as saying there are a few ways to think about "only if". Take the classical example of "If Socrates is a man, then he is mortal" to stand for $\displaystyle A \Rightarrow B$. This is equivalent to saying the following:
(1) "Socrates is a man, only if he is mortal." (A only if B)
(2) "If Socrates is not mortal, then he is not a man." (¬B implies ¬A)
(3) "Socrates is a man is a sufficient condition for his being mortal." (A is sufficient for B)
(4) "Socrates is mortal is a necessary condition for his being a man." (B is necessary for A)
The confusion people have is to think that whenever we see "if" what follows is the antecedent (A), but what follows "only if" is always the consequent (B). How does the "only" modifier change this, you might ask? I think of relating (1) with (2) and (4), which (2) really gets its force by thinking of it in terms of (4). I think most people confuse the "only" with making the "if" stronger, but then thinking the "if" plays its normal role (i.e., what follows is the antecendent). We also think of necessity in similar fashion. This is why I think of (1) in terms of (4). In short form, I think of it like this: A only if B is necessary for A. Therefore, the "only" modifies the if to be strong like a necessary statement, but not in terms of making it a definite $\displaystyle B \Rightarrow A$, which is wrong. It modifies it by exemplifying the necessity of B for A. I don't know if this helps, but it certainly helped me wrap my head around it.
Now, if we're concerned with just rules for remembering its form, I throw in the equivalence statement "iff" which stands for "if, and only if." The reason I hope is obvious. If we say "A if, and only if, B" then we're saying both $\displaystyle A \Rightarrow B$ and $\displaystyle B \Rightarrow A$. Since we know what follows the "if" is the antecedent, the "if" in "iff" applies to the second case. The "only if", therefore, applies to the first case. This is as expected: i.e., the first case is "A only if B" and the second case of the equivalence is "A if B".
I do not think that you will find the notation $\displaystyle A\not\Rightarrow B$ in a standard logic text books. (Mind you, I did not say that there is no textbook with that notation.)
As a matter of fact, both Irving Copi and Willard Quine take pains to point out that
$\displaystyle A\Rightarrow B$ is just an abbreviation for
$\displaystyle \neg \left( {A \wedge \neg B} \right)$.
Quine’s famous example is that “If anything is a vertebrate, it has a heart.” is a brief for “Nothing is a vertebrate and yet does not have a heart.”
Now consider this construction again: $\displaystyle Q\text{ is }\neg \left( {A \wedge \neg B} \right)$
Suppose that we know positively that B is TRUE.
Do you see that no matter what statement $\displaystyle A$ is we know that $\displaystyle Q\text{ is TRUE}!$.
Simply put: A true statement is implied by any statement.
for A and B, i like to use predicates that i understand well.
suppose A = "k is an integer which is a multiple of 6", and that B = "k is an even integer".
now, think of the following two statements: which one is true?
1) k is an integer which is a multiple of 6, only if k is an even integer.
2) k is an even integer, only if k is an integer which is a multiple of 6.
now think of the following two sets: X = {0,6,-6,12,-12,18,-18,.....} and Y = {0,2,-2,4,-4,6,-6,8,-8,......}
(i am being a bit imprecise, but X is integral multiples of 6, and Y is even integers).
which set is bigger? isn't it true that being in X implies being in Y? that is, n is in X ONLY IF n is in Y, because Y contains X (as a subset).
on the other hand, if n is NOT in Y (for example, if n is an odd integer), then we don't even have to check whether or not it's in X
(X is "totally inside Y". if we're "outside of Y", we can't get to something "inside of Y"). so NOT in Y, implies NOT in X (draw a picture).
the "flip side" of this coin, is that "A if B" means (somewhat confusingly) if B, then A:
3) k is an even number, if k is a multiple of 6 (that makes sense, no? all the multiples of 6 are even numbers).
our "natural language" is a bit ambiguous in this respect. consider the two statements (logically equivalent):
a) i'll kiss you if you let me.
b) if you let me kiss you, i will.
there is a subtle difference between those and:
c) i'll kiss you, but only if you let me.
in the first two, if the other person does not agree, they may get kissed anyway. in the last statement, if the other person does not assent, no kissing is going to transpire.