
Why is $p\wedge q\Rightarrow r$ is true, when p is true and q is false?
Answer
408.3k+ views
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.
Complete step by step answer:
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,
In this question, we are given that the predicted p is true and the predicted q is 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.
Complete step by step answer:
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 |
In this question, we are given that the predicted p is true and the predicted q is 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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

The correct order of melting point of 14th group elements class 11 chemistry CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE
