Validity of an implication based on a given implication

Question

If the proposition $\lnot p \to q$ is true, then the truth value of the proposition $\lnot p \lor \left ( p \to q \right )$, where $\lnot$ is negation,$\lor$ is inclusive OR and $\to$ is implication, is

• A. True
• B. Multiple Values
• C. False
• D. Cannot be determined

Answer 1

$$\sim p\rightarrow q$$ = p+q, and it is given that $$\sim p\rightarrow q$$ is true.

$$\sim p\rightarrow q$$ will be TRUE when one of the variable is true. So the valid value pair for p and q are (1,1), (1,0), and (0,1).

If we use these value pairs for $$\sim p\vee \left ( p\rightarrow q \right )$$ $\cong$ $\sim p \vee q$ we get final output as below

p	p'	q	p'+q
1	0	1	1
1	0	0	0
0	1	1	1

Here at the ouput column we are getting multiple values.

So, the answer is D.

Answer 2

I did this in the following way :

Truth value can be determined means : if A always 1 then truth then truth value of A is 1. Or,if A ia always 0 then truth then truth value of A is 0.

From the question valid combinations of p and q are restricted to (0,1) , (1,0) and (1,1).

Based on that, Truth value of last column is not always 1 or always 0.

=> can not be determined.

Comments

Should we say this as cannot be determined (or) Multiple values.

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Proposition definition itself says, either it is true or false but not both. So we can not say it will have multiple values, only we can say, it can not be determined..

Answer 3

given ¬p→q =True

(p')'+q=True => p+q=T

p	q	p+q
0	0	0
0	1	1
1	0	1
1	1	1

now check for

¬p∨(p→q)=p'+(p'+q)=p'+q

p'	q	p'+q
1	0	1
1	1	1
0	0	0
0	1	1

p'+q=p+q hence answe isTRUE 

please correct me if i wrong

Comments

HOW ? p'+q=p+q hence answe isTRUE.

p'+q and p+q have different truth table as well as different exp.

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according to TT

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wrong !!

You are comaparing two different truth table, look at your inputs, in the second TT, one of them is incomplement form.