Question
Question: If \(\left( p\wedge \sim q \right)\wedge \left( p\wedge r \right)\to \sim p\vee q\) is false , then ...
If (p∧∼q)∧(p∧r)→∼p∨q is false , then the truth values of p, q and r are respectively:
(A)F,T,F(B)T,F,T(C)F,F,F(D)T,T,T
Solution
For this problem, first we have to make a truth table for(p∧∼q) after that for (p∧r)
And at last for →∼p∨q, then we have to observe from the table the truth value of p and after that q and r.
Complete step-by-step solution:
We have to find the truth value of p, q, r when the (p∧∼q)∧(p∧r)→∼p∨q is false.
Now we will have to make the truth table with the help of its operation
| a | b | a→b | a∨b | a∧b | ∼a | ∼b |
|---|---|---|---|---|---|---|
| T | T | T | T | T | F | F |
| T | F | F | T | F | F | T |
| F | T | T | T | F | T | F |
| F | F | T | F | F | T | T |
With the using of the above table we can make (p∧∼q)∧(p∧r)→∼p∨q truth table:
| p | q | r | ∼p | ∼q | (p∧∼q) | p∧r | (p∧∼q)∧(p∧r) | ∼p∨q | (p∧∼q)∧(p∧r)→∼p∨q |
|---|---|---|---|---|---|---|---|---|---|
| T | T | T | F | F | F | T | F | T | T |
| T | T | F | F | F | F | F | F | T | T |
| T | F | T | F | T | T | T | T | F | F |
| T | F | F | F | T | T | F | F | F | T |
| F | T | T | T | F | F | F | F | T | T |
| F | T | F | T | F | F | F | F | T | T |
| F | F | T | T | T | F | F | F | T | T |
| F | F | F | T | T | F | F | F | T | T |
It is given that (p∧∼q)∧(p∧r)→∼p∨q is false. From the above table we can say that it's possible when p is true , q is false and r is also true.
Option B is the correct option.
Note: The symbol ∧ is used for and. Symbol ∨ is used for or. Symbol∼ is used for not.
Sometimes students are confused between a∧banda∨b and they make mistakes. Both the symbols have different meanings, if we use ∧ symbol instead of ∨ then our truth table is wrong.