Implication Operation $\rightarrow$

Question

Apologies in advance if it is stupid question, but I am learning it myself which means I may not have the prerequisites knowledge. please help.

I am reading "Introduction to the theory of computation (third edition)" by Michael Sipser. On page 15, on the second paragraph, "Finally, the implication operation is designed by the symbol $\rightarrow$ and its 0 if its first operand is 1 and its second operand is 0; otherwise, $\rightarrow$ is 1."

I understand what it means. I don't see how / where in math we would do that? the boolean is 0 if its first operand is 1 and its second operand is 0. Can you please one example in math / computer science ? Thank you so much in advance.

Answer 1

You should know that there are many students like you who have difficulty with material (logical) implication. How, for example, can we infer that $A\implies B$ is true knowing only that $A$ is false? Or knowing only that $B$ is true. In fact, we can make such inferences. These are well accepted methods of proof in mathematics or wherever we are dealing with logical propositions that are known to be unambiguously either true or false. It seems to be inevitable. It's not just some arbitrary convention that you must simply accept and memorize.

You may find my recent blog posting on material implication to be useful. There, I derive or attempt to justify each entry in the truth table for implication.

Answer 2

The conditional operation, $p\to q$, is $1$ if the first operand ($p$) is $0$, or if the first is $1$ and the second operand ($q$) is $1$.   The operation is only designated $0$ if the first operand is $1$ while the second is $0$.

This is summarised by the following "truth table": $${\begin{array}{|c:c|c|}\hline ~p~ & ~q~ & p\to q\\[0.5ex]\hline 1 & 1 & 1\\[0.5ex]\hdashline 1 & 0 & 0 \\[0.5ex] \hdashline 0 & 1 & 1 \\[0.5ex] \hdashline 0 & 0 & 1 \\[0.5ex]\hline \end{array}}$$

And why not?   We want the symbol to capture the notion that $p$ implies $q$.   That it is a promise that $q$ is $1$ when $p$ is $1$.   Such a 'promise' is only not held up when $p$ is $1$ but $q$ is not.   So that is the only row in the truth table we mark the operation with a value of $0$ .