This question already has an answer here:

$P_n$ and $Q_n$ are propositions which has a truth value $n$ where $n \in (T,F)$

So i want to access the truth value of $P\Rightarrow Q$


$P_T$ and $Q_T$, $(P \Rightarrow Q)_T$

$P_T$ and $Q_F$, $(P \Rightarrow Q)_F$

$P_F$ and $Q_T$, $(P \Rightarrow Q)_T$

$P_F$ and $Q_F$, $(P \Rightarrow Q)_T$

I don't really understand the last 2 statements. Can anyone explain them?

Like for (2), If P is true then Q is false, then $P \Rightarrow Q$ is false. that is self evidential. But how about (3) and (4) i can't seem to understand


marked as duplicate by MJD, Community Sep 6 '15 at 8:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ About the truth-functional definition of the connective $\to$ ("if __, then_"), you can see this post $\endgroup$ – Mauro ALLEGRANZA Sep 6 '15 at 7:57
  • $\begingroup$ And a bunch of other posts linked from that one. This is a FAQ. For example, Truth Table for If P then Q may be useful. $\endgroup$ – MJD Sep 6 '15 at 8:21
  • $\begingroup$ See also Peter Smith's hand-out. $\endgroup$ – Mauro ALLEGRANZA Sep 6 '15 at 9:32

In propositional calculus the "implies" connector means something a little more technical than implication in the cause and effect sense. It is usually the case that $A\implies B$ is shorthand for $\neg A \lor B$.


It's a definition.

$(P \Rightarrow Q)$ means that either (1), (3) or (4) is true.

In other words, $(P \Rightarrow Q) \leftrightarrow [(P \wedge Q) \vee(\neg P \wedge Q)\vee (\neg P \wedge \neg Q)]$

As (2), or $(P \wedge \neg Q)$, make (1), (3) and (4) all false, (2) also makes $(P \Rightarrow Q)$ false.


Not the answer you're looking for? Browse other questions tagged or ask your own question.