One thing that often confuses beginners in logic is that
$P\Rightarrow Q$ is TRUE when $P$ is FALSE, whatever Q is.
One (weak) reason is that we are mostly interested in the case where $P$ is TRUE, and with the above property, we have the nice formula $P\Rightarrow Q=(\textrm{non }P)\textrm{ or }Q$.
I am trying to justify the property by an axiomatic approach (to avoid to define $P\Rightarrow Q=(\textrm{non }P)\textrm{ or }Q$). For convenience, I will use $T$ for TRUE and $F$ for FALSE. Defining $P\Rightarrow Q$ means choosing a result in $\{T,F\}$ for all 4 possibilities of $P$ and $Q$ in $\{T,F\}$. So, I will use the functional notation $c(P,Q)$ to denote the resulf of $P\Rightarrow Q$. Clearly, if we want to express that $Q$ is the consequence of $P$, we should have the axioms:
- $c(T,T)=T$: if $P$ is TRUE, then $Q$ is also automatically TRUE.
- $c(T,F)=F$: $Q$ cannot be FALSE when $P$ is TRUE.
In addition, I would add
- Contraposition Principle: $c(P,Q)=c(\textrm{non }Q,\textrm{non }P)$
- Equivalence Axiom: $P\Leftrightarrow Q =(P\Rightarrow Q)\textrm{ and }(Q\Rightarrow P)$
where non, and and $\Leftrightarrow$ have their usual definition. Writing $a=c(F,T)$ and $b=c(F,F)$, I obtain the following truth table:
$$ \begin{array}{cccccc} P & Q & c(P,Q) & c(Q,P) & c(P,Q)\textrm{ and }c(Q,P) & P\Leftrightarrow Q \\ T & T & T & T & T & T \\ T & F & F & a & F & F \\ F & T & a & F & F & F \\ F & F & b & b & b & T \\ \end{array} $$
so the Equivalence Axiom forces $b=T$. The Contraposition Axiom would have the same consequence.
Question: My axioms say nothing about $a$. What could be a natural axiom to add to the list to obtain $a=T$?