# On the truth-value of implication connective

As I have come to understand, in classical logic, the implication statement turns out to be true if the premise is false. It seems to be a little counter-intuitive, as it seems to me that the truth value should ideally be neither true nor false . Is there any other logic system (maybe some multivalued logics ?) where such a stance is taken ?

EDIT : I've encountered the promise-analogy quite a number of times, but that too seems to be dis-satisfactory - since it implicitly assumes a 2-valued logic - whereas it seems fitting that a third value (maybe something like "undefined") be accommodated into the system. I don't know, but there maybe some technicalities involved in such a thing. It would be helpful if someone could explain.

You can see in this post the beautiful Henning Makholm's answer :

However, one should note that these [the "usual"] arguments are ultimately not the reason why ⇒ has the truth table it has. The real reason is because that truth table is the definition of ⇒. Expressing p⇒q as "If p, then q" is not a definition of ⇒, but an explanation of how the words "if" and "then" are used by mathematicians, given that one already knows how ⇒ works. The intuitive explanations are supposed to convince you (or not) that it is reasonable to use those two English words to speak about logical implication, not that logical implication ought to work that way in the first place.

I completely agree with him; my personal understanding of this issue is in the answer to this post.

I hope it can help...

With natural deduction for classical logic, we can derive the equivalence between $A \rightarrow B$ and $\lnot (A \land \lnot B)$.
For an implication, it is clear that $v(A \rightarrow B) =$ f if $v(A) =$ t and $v(B) =$ f. For the remaining three cases, we can use the classical equivalence of $A \rightarrow B$ and $¬(A \land ¬B)$ : If $v(¬(A \land ¬B)) =$ t, then $v(A \land ¬B) =$ f by [...] the value of negation. Then either $v(A) =$ f or $v(¬B) =$ f. In the latter case, $v(B) =$ t. Thus, an implication $A \rightarrow B$ has the value t under a given valuation if either $A$ has the value f or $B$ has the value t, otherwise $A \rightarrow B$ has the value f.