I understand P ∧ Q, being that both must be equivalent, ie True & True, or False and False.

I understand P --> Q implies that if P is True we know what Q is and if Q is true then the result is True, if P is True and Q is False then the result is False, if P is False we do not know what Q is and assume it to be true.

So for (P ∧ Q) --> R, does this mean I do P && Q first and use those results as P as though I was doing P --> Q and R being Q.

I used a calculator online and it showed me this as the answer but I didn't understand it:

p   q   r   ((p ∧ q) → r)
F   F   F       T
F   F   T       T
F   T   F       T
F   T   T       T
T   F   F       T
T   F   T       T
T   T   F       F
T   T   T       T

So looking at the first row, p=F and q=F therefore p∧q is True?, r is False therefore the last column should be False no? Why is it True?

Or is the result true only when both are in fact True, not just equivalent? F ∧ F = F?

  • 4
    $\begingroup$ For $p \wedge q$ to be true, both $p$ and $q$ must be true. If either of them is false, then $p \wedge q$ is false. $\endgroup$ – Aniruddha Deshmukh May 11 at 7:51
  • 1
    $\begingroup$ "I understand $P ∧ Q$, being that both must be equivalent, ie True & True, or False and False." NO : $\land$ means "and" and thus $P ∧ Q$ is true when both $P$ and $Q$ are true $\endgroup$ – Mauro ALLEGRANZA May 11 at 13:27

Some overarching ideas:

  • You could treat $P \land Q$ as a single statement.
    • Thus for example you could look at $S \implies R$. Determine the truth value table for that.
    • From there you can expand said table by looking at $S$'s truth values when $S$ is $P \land Q$.
    • $P \land Q$ is only true if both $P,Q$ are.
  • $X \implies Y$ does not require that $X,Y$ are both true, just that a false implication does not follow from truth.
    • That is to say, truth implies truth. This, $X$ true and $Y$ true give $X \implies Y$ truth.
    • $X$ false implies nothing about $Y$ necessarily, so $Y$ could be true or false. Implication is predicated on the precedent being true. Thus if $X$ is false, $X \implies Y$ is true outright.
    • If $X$ is true and $Y$ is false, then obviously $X \implies Y$ is untrue.

Let $S$ represent the truth value for $P \land Q$ under this premise then. Then for the statement $S \implies R$ we get the usual truth table:

$$\begin{array}{cc|c} S & R & S \implies R \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array}$$

What's the truth table for $S$, i.e. $P \land Q$?

$$\begin{array}{cc|c} P & Q & P \land Q \\ \hline T & T & T \\ T & F & F \\ F & T & F \\ F & F & F \end{array}$$

Combining the two, then,

$$\begin{array}{cc|c|c|c} P & Q & S = P \land Q & R & S = P \land Q \implies R \\ \hline T & T & T & T & T \\ F & T & F & T & T \\ T & F & F & T & T \\ F & F & F & T & T \\ T & T & T & F & F \\ F & T & F & F & T \\ T & F & F & F & T \\ F & F & F & F & T \end{array}$$

In summary:

Whenever the precedent $P \land Q$ is true but $R$ is false is the only case when the implication $P \land Q \implies R$ is false, because of how implication works from a logical perspective (it's reliant only on the truth of the precedent). In all other cases, we claim the implication to be true, since either the implication holds or the precedent is false and so has no bearing on the implication.

From there, we consider the circumstances under which the precedent $P \land Q$ is false or true, consider its effects on the implication, and make our truth table from there.


The only way to falsify $(p \land q) \to r$ is to make $p \land q$ true and $r$ false. So in the truth table there will be all true except when $p,q$ are true and $r$ false.


Maybe the problem comes from the fact you do not recognize the structure of the statement due to the fact that the antecedent is not an atomic formula.

Your main statement is a conditional.

The structure of a conditional is always :


The internal structure of the antecedent ( considered alone) or of the consequent ( considered alone) never affects the structure of the whole conditional, nor its truth-conditions.

The only requirement is that both " antecedent" and " consequent" be propositions, well formed formulas.

A proposition, a well formed formula, need not be an atomic sentence ( though it also can be an atomic one).

For example the whole biconditional formula : ( A <--> ( (B v ~ C) --> D ) )

can perfectly play the role of antecedent or of consequent in a conditional statement.

The method to determine the possible truth values of the whole conditional is always the same ( the only changes from one conditional to another will be the number of lines in your truth table, depending on the number of atomic sentences involved in the antecedent and the consequent):

(1) compute the possible truth values of the whole antecedent ( however complex the formula may be)

(2) compute the possible truth values of the consequent ( however complex...)

(3) compute the possible truth values of the whole conditional using the truth-table of the " if .. then " operator.


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