# In propositional logic how is the negation equivalent?

I am unable to understand how the negation of a sentence into CNF actually makes sense, hopefully somebody can explain it to me in understandable terms!

Given the sentence: If the battery is charged (A) and the screen is not cracked (B) then the phone is working (C)

The following makes perfect sense:

(A ∧ B) => C

I cannot get my head around how the negated form is true or how it works:

¬A ∨ ¬B ∨ C

I am struggling because "not A" means if the battery is not charged? So I cannot see how if the battery is not charged or the "screen is cracked" or the "phone is usable".

If A is false, so battery is not charged, how is that applying ¬A to it does not make it now true?

Any help would be appreciated.

• I guess your negated formula is wrong? How do you get $\neg B \lor C$ from $B \implies C$? – Algebruh Feb 1 at 8:33
• Maybe it is, can you explain a little more about what you mean? I took it to mean "its not charged, or the screen is cracked, or it works" – pac234 Feb 1 at 21:00
• I missed something out...so nevermind :) – Algebruh Feb 1 at 21:40

First thing is to accept and understand, that $$A \Rightarrow C$$ is same as $$(\neg A) \lor C$$. Using your example we obtain, that "If the battery is charged, then the phone is working" is same with "the battery is not charged or the phone is working".

Now following initial sentence $$(A \land B) \Rightarrow C$$ we have $$\neg(A \land B) \lor C$$, where last sounds as "not fulfilling (the battery is charged and the screen is not cracked) or (the phone is working)".

Now we come to negation of $$\neg(A \land B) = (\neg A) \lor (\neg B)$$. So sentence "not fulfilling (the battery is charged and the screen is not cracked)" is same as "(the battery is not charged) or (the screen is cracked)"

Putting everything together we have $$(A \land B) \Rightarrow C \text{ is same with } (\neg A) \lor (\neg B) \lor C$$

in words last is "(the battery is not charged) or (the screen is cracked) or (the phone is working)"

• Thanks this makes a lot of sense and is what I was missing. The only thing still confusing me now is how the OR's are executed, so it is not to be read like a set of OR statements inside an if condition, more like an if else (the final OR being the else)? Or should I not try to liken this to if conditions? – pac234 Feb 1 at 8:46
• "then .. if.." becomes, or more exactly, is "not .. or.." - this is one of the most confusing moments in mathematical implication. More human may be sounds "or not .. or..". One example: "if I have money, then I'll buy book" is same with "or I haven't money or I'll buy book". So implication is not set of "or"s inside "if", but "if" itself is "or not .. or..". Continuing.. – zkutch Feb 1 at 8:56
• Continuation. If you look more closer, on some next step in future, then you'll see, that in truth table such definition of implication gives "strange" behaviour when premise of implication is false: despite of what conclusion is - true or false - whole implication become true. – zkutch Feb 1 at 8:57
• Ok so its much more like a human readable sentence in theory – pac234 Feb 1 at 9:05

$$\neg A \vee \neg B \vee C$$ means that the situation: "the battery is charged", "screen is not cracked", and "phone is not usable", is FALSE or not possible. The rest of the possibilities are all true.

This is equivalent to say that when "the battery is charged" and "screen is not cracked", the phone must be usable, i.e., $$(A \wedge B) \to C$$