Why $F \rightarrow T \implies \neg F \rightarrow \neg T \implies T \rightarrow F$ is wrong?

Question

$$ F \rightarrow T \implies \neg F \rightarrow \neg T \implies T \rightarrow F $$ Why is this reasoning wrong according to the truth table of implication in logic?

Thanks.

Answer 1

Let's go through your statements one at a time: $$ F \to T $$ Looking at the truth table, this evaluates to $T$ (true). $$ \lnot F \to \lnot T $$ (I use $\lnot$ to denote "not") -- this one simplifies to $T \to F$. Then looking at the truth table, it's $F$. $$ T \to F $$ As we just saw, that is also $F$ (false).

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So it looks like your problem is going from $F \to T$ to $\lnot F \to \lnot T$. That jump is not correct. In general, $$ A \to B \text{ is equivalent to } \lnot B \to \lnot A $$ but $$ A \to B \text{ is not necessarily equivalent to } \lnot A \to \lnot B. $$

Answer 2

This is because, unlike algebra with equality operations $=$, you do not preserve truth in logical implications $\rightarrow$ by performing the same operation on both sides.

I assume you started with $$ F \rightarrow T $$ which is correct, and then proceeded to negate both sides like this $$ \neg F \rightarrow \neg T $$ However this is not true because of how $\rightarrow$ works with the implication.