I am struggling trying to get a propositional formula of the following truth table: A, B, AC, BC.
For A and B it's easy: A xor B. However, when you insert a new literal C...
I tried using Wolfram by inputting the truth table (A & ~B & ~C) || (~A & B & ~C) || (A & ~B & C) || (~A & B & C). However, the suggested minimal forms are wrong since they do not consider C.
Can someone help expressing this in propositional logic using logical connectives like (A xor B) => C ? Thanks!
You can perform minimization by means of using Karnaugh maps (amongst other methods - this one is the simpliest, you'll have to introduce a dummy variable D and just ignore it in the results).
The solutions are right about not considering C
, though - it doesn't matter what the C
evaluates to, as long as A xor B
evaluates to true. I just did check that, to remind myself about how the Karnaugh maps are constructed. Try drawing yourself a full truth table to see that.
Take a look at the expression:
(A & ~B & ) || (~A & B & ) ||
(A & ~B & ) || (~A & B & )
Both lines are identical except for the negation of C
, this means that C
is irrelevant : the value for C
doesn't change anything to the output of the function.
This is also the conclusion one draws from a truth table:
|A|B|C||F|
+-+-+-++-+
|F|F|F||F|
|F|F|T||F|
|F|T|F||T|
|F|T|T||T|
|T|F|F||F|
|T|F|T||F|
|T|T|F||F|
|T|T|T||F|
Here F
being the outcome of the expression. If you for instance take the first line: |F|F|F||F|
the result is false, which is the same for |F|F|T||F|
(with C
flipped). By doing this for every ( A
, B
) configuration one sees that the value of C
doesn't matter.
Therefore you can simply exclude C
from the formula resulting in:
(A & ~B) || (~A & B)
Which means A xor B
.
Wolfram Alpha comes to the same conclusion (see ANF expression).
I have the answer
(A xor B) and (C => (A or B))
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