Motivating the Implication Operator

Question

I am self-studying from Number Systems and the Foundations of Analysis By Eliott Mendelson.

The book has just defined the truth table for the implication operator and in the following problems seeks to motivate why everything other than T$\Rightarrow$F is evaluated as T. The question begins as follows:

"Verify the following justification for our truth table for A$\Rightarrow$B. It is obvious that $(C\land D)\Rightarrow C$ must always be true. Hence: (i) When $C$ and $D$ are both true, we have $(T\Rightarrow T)=T$. This gives the first row of the table for $\Rightarrow$. (ii) When $C$ is false and $D$ is true, we have $(F\Rightarrow T)=T$, which is the second line of the table. (iii) When $D$ is false, we have $(F\Rightarrow F)=T$, which is the fourth line of the table."

$$\begin{array}{ccc} A & B & A\Rightarrow B \\ \hline T & T & T\\ F & T & T\\ T & F & F\\ F & F & T \end{array}$$

My question is what about the assertion $(C\land D)\Rightarrow C$ is obvious? You can of course use the truth table to see that this is true. But if the question is trying to motivate the truth table, I'm left to assume that I can't use the truth table. So it must be true on the basis of an intuitive understanding of implication, or something else.

Can anyone shed light on why this would be an "obvious" assertion?

Answer 1

Consider this statement:

"If I have a car and a donkey, then I have a car"

I assume you would consider that true (and all very logical), right?

Well, consider symbolizing as follows:

$C$: I have a car

$D$: I have a donkey

Then the above sentence symbolizes as: $(C \land D) \to C$

I hope you can see that this always works, no matter what $C$ and $D$ are: if $C$ and $D$ are both true, then $C$ should be true.

Of course, I am not saying that $C \land D$ is true. I am also not sdaying that $C$ is true. But I am saying that $C \land D$ logically implies $C$. That is: in any situation where $C \land D$ is true, we find that $C$ is true. Or: there is not situation where we finds that $C \land D$ is true, but $C$ is false: Can you imagine a world where I have both a car and a donkey, but I don't have a car?

So: $C \land D$ logically implies $C$. But if that is the case, then $C \land D$ should always materially imply $C$ as well. Hence, $(C \land D) \to C$ should always be true.

Remember that the logical connectives that we mathematically define are meant to capture certain truth-functional relationships. The $\land$, for example, is mean to try and capture what we mean by the English 'and'. When we say 'C and D', we are claiming that both C and D are true. You see this reflected in the truth-table. So it is not that we say that "C and D" is true whenever $C$ and $D$ are true because the truth-table says so, but rather it is just thew other way around: we define the truth-table in a way so as to reflect what we mean by 'and'.

Likewise, we try to capture the truth-functional relationship expressed by any English 'if ... then ...' statement by mathematically defining a $\to$ truth-functional operator. Now, our conception of the 'if ... then ...' is such that when we say 'If both $C$ and $D$ are true, then in particular $C$ is true by itself', we expect to have just stated a true 'if ... then ...' claim. And, we feel that this should be true no matter the truths of $C$ and $D$. If I say 'If I have a car and a donkey, then I have a car', then any 'normal' speaker of the English language will say : "sure, makes sense".

And does it make sense only when in fact I do have a car and a donkey? No. The statement also makes sense if I don't have a car or a donkey. It would still make sense to say that if I have a car and a donkey, then I would have a car.

And again, we don't say that "if I have a car and a donkey, then I have a car." is true because of how the $\to$ is mathematically defined. Rather, it is just the other way around: the $\to$ is mathematically defined the way that it is because of arguments like the one provided in your textbook: because speakers of the English language regard "if I have a car and a donkey, then I have a car" to always be true.

Answer 2

About the implication connector,

to avoid the cases where the hypothesis is false, and we say that the implication is true,

I purpose a new truth table for the implication .

$(P \implies Q) \;\; P \;\; Q$

$ T \;\;\;\;\;\;\;\;\;\;\;\; \;\;\; T. \;\;\;\; T$

$ T \;\;\;\; \;\;\;\;\; \;\;\;\; F. \;\;\;\; ?$

$ F. \; \;\;\;\;\;\;\;\; \;\;\; T. \;\;\;\; ?$

$ F. \;\;\;\;\;\;\; \;\;\;\; F. \;\;\;\; ?$

Answer 3

We do not take implication as some form of causality, but as a form of logical consequence, i.e., that the truth of a first thing forces us to accept that a second thing is true. The essence of "and" is: that one thing and another are true necessitates that each of these two things is true. So with $C\land D$ representing the statement that $C$ and $D$ are true forces us to accept that $C$ is true (it also forces us to accept that $D$ is true). If you do not agree, you may have an unusual notion of "and" (or of implication). And this fact, i.e., that the truth of $C\land D$ forces us to accept that $C$ is true. The accepting-force behind this argument is always correct, no matter whether $C\land D$ itself is actually true. This now precisely what $(C\land D)\Rightarrow C$ symbolizes.

Answer 4

A more standard truth table:

I don't see what $C\land D \implies C$ has to do with it, but the truth table for IMPLIES can be justified as follows using a form of natural deduction:

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