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# Why is $p\wedge q\Rightarrow r$ is true, when p is true and q is false?

Last updated date: 23rd Jul 2024
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Hint: We know that if we have two predictions p and q, then the implication $p\Rightarrow q$ is true when either p is false, or q is true, or both. Also, we must know that a conjunction is true only when both the predictions are true. Using these two concepts, we can show that $p\wedge q\Rightarrow r$ is true when p is true and q is false.

We know that if we have two predictions p and q, then $p\Rightarrow q$ is true when either p is false, or q is true, or both.
We also know very well that the conjunction $p\wedge q$ is true only if both p and q are true, else the conjunction is false. We can also see this from the truth table shown below,
 p q $p\wedge q$ True True True True False False False True False False False False
From the truth table, we can see that when p is true and q is false, then the conjunction $p\wedge q$ is false.
And from the above discussion on implication, we can say that the implication $p\wedge q\Rightarrow r$ is true when either $p\wedge q$ is false, or r is true, or both.
We have no information about r. Now, we have proved above that the conjunction $p\wedge q$ is false.
So, we can easily say that the implication $p\wedge q\Rightarrow r$ is true.
Hence, when p is true and q is false, the implication $p\wedge q\Rightarrow r$ is true.
Note: We must remember that an implication is also called a conditional. We should also know that the conjunction, represented by the symbol $\wedge$, is also called AND logic. We can also solve this problem using negation and disjunction.