edited by
360 views
2 votes
2 votes
Let $\text{P}$ be a compound proposition over $4$ propositional variables $: a,b,c,d.$

We know that for $a$ compound proposition over $n$ propositional variables, we have $2^{n}$ rows in the truth table.

Every row of the truth table of $\text{P}$ is called an “Interpretation” of $\text{P}.$

A row in the truth table of $\text{P}$ is called “model” iff $\text{P}$ is true for that row.

Let $\text{P}$ be “ $a \leftrightarrow b$”

How many models are there for $\text{P}?$
edited by

2 Answers

Best answer
3 votes
3 votes

 Given P is “ a↔b ”,

P becomes true only when (a=0 and  b=0) or (a=1 and b=1) so here we have two choices.

for the remaining variables(c,d), we can choose 0 or 1(as it doesn’t affect P). so we have 2*2=4 choices.

The total no of models are 2*4=8.

edited by
Answer:

Related questions

3 votes
3 votes
2 answers
4
GO Classes asked Apr 14, 2022
360 views
Consider the following statement $\text{S}$ in an universe $\text{U}.$$\text{S} : \forall x \forall y (x = y)$What is the maximum cardinality of $\text{U}$ such that $\te...