Number of rows which will evaluate to be true

Question

Question

Consider the following well defined formula -:

$$(p \vee q'\vee r' \vee s) \rightarrow t \vee u'$$

The maximum number of rows in truth table of above formula which evaluates to be true is?

My Approach

\begin{align}\text{Let } Q &=(p \vee q' \vee r' \vee s) \rightarrow t \vee u' \\ &=(p \vee q' \vee r' \vee s)' + t + u' \\ &=p'qrs' + t + u' \end{align}

$Q$ will always evaluates to be true except when all $p'qrs',t,u'$ will be false

I posted this question here because I think it as a question of combinatorics.

So the number of rows which will evaluate to be true=$$2^{3}-1=7$$

Am I correct?

Answer 1

HINT

There are $6$ variables, so $2^6=64$ rows

Any conditional is true iff either the antecedent is false or the consequent is true.

So, find all rows where the consequent $t \lor u'$ is true (this will be the case in $3$ out of $4$ cases, so out of the $64$, $48$ rows are like that)

Then find all the rows where the antecedent is false: this will be the case in $1$ pout of $16$ rows, i.e. out of $64$ rows, there are $4$ rows like that.

OK, so that seems to give $48 + 4 = 52$ rows .. but did you double count? That is, in how many rows is the antecedent false and the consequent true? That number will need to be subtracted ... I'll leave that to you to figure out