Since $F \Rightarrow F$ and $F \Rightarrow T$ both evaluate to $T$ with the truth table for $\Rightarrow$, does this not break mathematical induction?
For example, once you show the base case holds for a proposition $P$, then you could do the induction hypothesis as follows: "Suppose $P(k)$ does not hold. Since $P(k+1)$ will either hold or not with this assumption, $P(k) \Rightarrow P(k+1)$, thus $P(n)$ holds for all $n \in \mathbb{N}$."
Answer
No, it doesn't. You can only deduce from true statements. That's independent of the truth table (and even is true in logic systems that don't know truth tables).
But imagine if $F\implies T$ would be false. Then you could prove $1=0$ as follows:
Let's assume $0=1$ were false. Now $0=1 \implies 1=0$ (symmetry). But $1=0\land 0=1\implies 1=1$ (transitivity). Therefore we have proven $0=1\implies 1=1$. But since $1=1$ is undeniably true, that would mean a false statement implies a true statement, violating that $(F\implies T)$ is false. Thus $0=1$ cannot be false by contradiction.
As you can see, actually $(F \implies T)=F$ would break logic.
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