# If $b$ is tautology and $c$ is contradiction why isn't $a\to (b\to c)$ a contradiction?

Let $a$ be neither tautology nor contradiction, $b$ is tautology and $c$ is contradiction . What is the value of $a\to (b\to c)$?

It's a multiple choice question where there're following options:

1. the expression is a contradiction

2. the expression is equivalent to the negation of $a$

The second answer is correct but I don't understand why. $b\to c$ is always false. Therefore $a\to (b\to c)$ is always false. So why isn't it a contradiction?

I derived my conclusion by using the truth table in this Wikipedia article.

$(b→c)$ is FALSE because - being $b$ a tautology, i.e. always TRUE, and $c$ a contradiction, i.e. always FALSE, it is :

(TRUE → FALSE), i.e. FALSE.

But $a$ can be either TRUE or FALSE.

Thus $a→(b→c)$ can be either (TRUE → FALSE) or (FALSE → FALSE).

More specifically, we have that :

$a→(b→c)$ is $a \to$ FALSE.

Thus, when $a$ is TRUE, we have that $(a \to$ FALSE) is FALSE, and when $a$ is FALSE, $(a \to$ FALSE) is TRUE.

Thus :

$a→(b→c)$ is equivalent to $\lnot a$.

$b\to c$ is always false. CORRECT
Therefore $a\to (b\to c)$ is always false. INCORRECT
$b \to c$ has its truth value specified because $b$ and $c$ each have their truth value specified. By definition $\top \to \bot$ is false, so you have $a \to \bot$ which is by definition equivalent to $\neg a$.