Logic consequence of a logic sequence

Question

Imagine:

A ⇒ B ⇒ C ⇒ D

You're asked to find whether the formula is logically true. Is in this case, D a logical consequence of C, which is a logical consequence of B, which is a logical consequence of A? So that you can prove that the formula is true by proving

A ⇒ B and B ⇒ C and C ⇒ D

Thank you for your time

Answer 1

$A \Rightarrow B \Rightarrow C \Rightarrow D$

is simply mathematical short-hand for:

$A \Rightarrow B$ and $B \Rightarrow C$ and $C \Rightarrow D$

so obviously if you can prove the latter, you will have proven the former, since it really is the same thing

Also, in the Comments you ask:

is it sufficient to prove that: A = B and B = C and C = D rather than rewriting all the implications as disjunctions and going from there?

Hmm, be very careful here! First, when you say $A = B$, do you mean that $A$ and $B$ are the exact same identical statements, or do you mean that $A$ and $B$ are logically equivalent statements? For example, $P \land Q$ and $Q \land P$ are logically equivalent, but they are not identical. You could say that maybe they are logically 'identical', but they are not syntactically identical

Now, it turns that, whether you mean logical equivalece or syntactically identical, if $A=B$, then certainly $A \Rightarrow B$, so yes, showing $A = B$ and $B = C$ and $C = D$ will indeed prove that $A \Rightarrow B \Rightarrow C \Rightarrow D$

However, please notice this: syntactical equivalence is obvious; either the statement is exactly the same statement or not. So, I am guessing that's not what you meant. I think you meant logical equivalence. Thus, for example, if $A$ is $\neg(P \land Q)$, and $B$ is $\neg P \lor \neg Q$ then, by DeMorgan's Laws, they are logically equivalent, and we can say that '$A = B$' .... though most logicians prefer to write this as $A \Leftrightarrow B$

OK, but now think about this: what would it take to show '$A = B$', i.e. $A \Leftrightarrow B$? In almost all practical scenarios, this is done by showing two things: that $A \Rightarrow B$, and that $B \Rightarrow A$. But wait! All you wanted to show is that $A \Rightarrow B$. So sure, showing that '$A = B$' will prove that $A \Rightarrow B$, but it makes little sense to show all that, if all you want to show is just $A \Rightarrow B$.

Also, when you say:

rather than rewriting all the implications as disjunctions and going from there?

you are making a horrible mistake: The $\Rightarrow$ implication is a logical implication, which is not the same as the $\to$ material implication. And yes, for the material implication it holds that $P \to Q \Leftrightarrow \neg P \lor Q$, but for the logical implication there is not such thing.

EDIT

I just realized that maybe you are dealing with one of those bad texts that use $\Rightarrow$ to represent the material implication. If so: Aaargh! OK, so if that is what is being meant, then we're not dealing with mathematical shorthand for a chain of logical implications, but with a single logic formula that would be: $A \to B \to C \to D$ ... except the problem with that is that the formula is ambiguous, as pointed out by @Stinking Bishop in the other Answer. So ... that makes me believe that we are not dealing with a single logic formula, but again, with a common mathematical short-hand notation for a chain of logical implications. Then again, you did state:

You're asked to find whether the formula is logically true.

and that suggests that maybe they did mean for this to be a logic formula ... even though it is ambiguous.

Answer 2

I am not convinced that the expression $A\implies B\implies C\implies D$ is unambiguous. Even though you can evaluate it left-to-right, I would find it confusing if you didn't put some brackets. So in what follows I've put brackets where I guess you would want them - feel free to put them elsewhere and conduct a similar analysis!

So, if you are asking whether $$((A\implies B)\implies C)\implies D$$ is equivalent to $$(A\implies B)\land(B\implies C)\land(C\implies D)$$ it can be easily shown that they are not equivalent, because the former is true if, for example, $A, C, D$ are true and $B$ is false, while the latter one is false in that case.

Answer 3

Whichever bracketing for $A\to B\to C\to D$ was implied the formula does not mean $A\to B$, $B\to C$, and $C\to D$. Consider the conventional way to read $A\to B\to C\to D$ as $A\to (B\to (C\to D))$. This formula only suggests that $A$ implies $B\to (C\to D)$. It doesn't (for example) say that $C$ implies $D$. The same logic applies to other ways of bracketing.

Here is an informal example to illustrate this point: Say if I feel good, then if it's not raining I'll go for a walk is true. This sentence has the form $A\to (B\to C)$, where

$A$ is I feel good

$B$ is it's not raining

$C$ is I'll go for a walk

Notice that if it's not raining I'll go for a walk a.k.a. $B\to C$ is not necessarily true (maybe I don't feel good). If I feel good, then it's not raining a.k.a $A\to B$ is also not necessarily true (it's not even a part of the formula!)

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To prove $A\to (B\to (C\to D))$ it would actually suffice to prove only $C\to D$. However, even if you prove $A\to B$, $B\to C$, and $C\to D$ it could be not the case that $(A\to (B\to C))\to D$. Indeed, consider the case when all $A, B, C$, and $D$ are false.

So the answer to

Is in this case, D a logical consequence of C, which is a logical consequence of B, which is a logical consequence of A?

is no.

The answer to

So that you can prove that the formula is true by proving A ⇒ B and B ⇒ C and C ⇒ D

depends on the bracketing.