In mathematical relations, then and and are represented by ->→ and ∧∧respectively.
So, we are to prove that either (P->Q)∧(P->R)(P→Q)∧(P→R) -=≡ P->(Q∧R)P→(Q∧R)
or (P->Q)∧(P->R)(P→Q)∧(P→R) cancel(-=) P->(Q∧R)
Note: I have represented equivalent sign with -= and not equivalent sign with cancel(-=).
To find that either two statements are logically equivalent or not we use truth table. Before finding the truth tables, let me tell you the mathematical definitions of then and and relations.
then: A then statement of two statements x and y, i.e, x->y is false only when x is true but y is false, otherwise, it is always true. A then statement is known as implication.
and: An and statement of two statements x and y, i.e, x∧y is true only when x and y both are true, otherwise, it is always false. An and statement is known as conjunction.
Keep above definitions in mind.
In your given statement there are three basic statements P, Q, R and five compound statements P->Q, P->R, Q∧R, (P->Q)∧(P->R), P->(Q∧R). So, there are eight statements in total.
The basic three statements give rise to 2^3=8 possibilities. Now, draw a table with 9 rows and 8 columns.
In the first row, list all the statements as shown below.
![enter image source here]()
Three basic statements give rise to 8 possibilities which can be listed in the table as shown below.
![enter image source here]()
Now, fourth column represents P->Q statement, this will be false only when P is true but Q is false. If you look into the table you will observe that P->Q is false only for 3rd and 5th row. Similarly the statement P->R, in 5th column, is false only for 2nd and 5th row.
![enter image source here]()
Now, sixth column represents Q∧R statement, this will be true only when both Q and R are true, otherwise false. In the table, Q andR both are true only in 1st and 4th row.
![enter image source here]()
For seventh column, you need to look in 4th and 5thcolumn and perform and operation on them, i.e, 7th column is conjunction of 4th and 5th column. Therefore, 7th column is true whenever both 4th and 5th columns are true. After looking into the table, I find that 7th column is false for 2nd, 3rd and 5th row.
![enter image source here]()
Finally for eighth column, you need to look into 1st and 6th column and perform then operation on them, i.e, 8th column is implication of 1st and 6th column. Therefore, 8th column is false only when 1st column is true, but 6th column is false. After looking into the table. I find that 8th column is false for 2nd, 3rd and 5th row.
![Truth Table]()
After very good hard work, the truth table is ready.
If you look into 7th and 8th column you will find that both the columns have same values for all the eight possibilities. That means that 7thand 8th columns are logically equivalent. Since, 7th and 8th columns represent (P->Q)∧(P->R) and P->(Q∧R) respectively, hence, they are logically equivalent.
implies (P->Q)∧(P->R) -= P->(Q∧R)
Hence, proved!
