2
$\begingroup$

Hi all this is my first time using this site so I hope I am presenting this properly, apologies if not. It is said that “a → b” and “¬(a and ¬ b)” are logically equivalent but I do not understand why/agree. However I agree that “a → b” and “¬a or b” are logically equivalent due to their truth tables being the same.

Here is why I do not understand (please point out where I made a mistake – I know I must have): The truth table for a → b is like this:

a   b   a → b

T   T   T
T   F   F
F   T   T
F   F   T       

I can understand this and am happy with it. However this is now where I do not agree with what is accepted. The truth table for ¬(a and ¬b):

a   b   ¬(a and ¬ b)

T   T   T
T   F   F
F   T  (F)
F   F   T       

Now I disagree on the third row and now I will show my working for that: ¬(a and ¬ b) in words is:

not(a and not b)        insert truth values for a and b
not(F and not T)        which simplifies to 
not(F and F)            which simplifies to 
not(T)                  which simplifies to 
F

This is why I think that it should be false on the third row which would disagree with the third row for a → b. I know I must have made a mistake/logical error but at the moment I cannot see it. I am new to studying logic so if anyone could point out what I have done wrong that would be great.

$\endgroup$
3
  • 1
    $\begingroup$ $F \land F$ is $F$, remember the truth table for and, you get $T$ only when your inputs are both $T$ and otherwise false. $\endgroup$
    – James
    Commented Jul 15, 2014 at 15:27
  • $\begingroup$ Thanks James I knew I made a stupid mistake. Cheers for the help. $\endgroup$
    – Mick
    Commented Jul 15, 2014 at 15:29
  • $\begingroup$ In the truth-table for ¬(a and ¬b), third row, you have written (F). Why ? (a and ¬b) in the case F-T is F-F which is F; thus its negation : ¬(a and ¬b) will be T, as expected. $\endgroup$ Commented Jul 15, 2014 at 15:30

1 Answer 1

2
$\begingroup$

Let's recall the truth table definition for $\land$ (and):

$T \land T = T$

$T \land F = F$

$F \land T = F$

$F \land F = F$

$A \land \lnot B$ under the assignment $A = F, B= T$ is $$F \land \lnot T = \;F \land F = F$$

That makes $\lnot (A \land \lnot B)$ evaluate to $\lnot (F) = T$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .