# Why can't three-valued logic (ternary logic) simply have only two truth values?

Consider the statement:

P ∧ ¬P ⊢ Q

where:

• P is any proposition. -¬P is the negation of P.
• Q is another proposition.

Wouldn't proving both P and ¬P to be true simply lead to a new proposition Q, rather than introducing a third truth value?

Even if we follow the principle of explosion, wouldn't the result still be either true or false, rather than a third truth value? This principle does not mean that P=¬P; rather, it is used to prove the truth of any other proposition Q, regardless of the content of P or ¬P.

For example, From P ∧ ¬P, one can deduce P ∨ Q (distribution rule) and from P ∨ Q, and because P is true (from P ∧ ¬P), Q (exclusion rule) can be deduced.

Thus, P=¬P does not exist on its own but is considered under the condition ∧ Therefore, the first proposition P is either true or false, and the third value is just a new proposition Q. How Can this new proposition be a third truth value without it being a proposition?

• If the system proofs both $P$ and not $P$ to be true , then it is inconsistent and therefore useless.Maybe , you mean the case that the system neither can prove $P$ nor not $P$. In this case, $P$ is independent of the system and we can either add $P$ or not $P$ as a new axiom without creating (new) contradictions. Commented Jun 15 at 9:56
• @Peter i meant in them being both true is for example "The sky is cloudy." There are three possible judgments: 1. True: The sky is entirely covered with clouds. 2. False: The sky is completely clear. 3. Intermediate: The sky is partially covered with clouds. "The sky is cloudy" is true to an intermediate degree here because there are clouds, but it is also partially false because the sky is not fully covered with clouds. So p and not p is true at the same time, but that would mean a indeterminate value, but wouldn't this value be a new proposition (Q)? And not a third value?
– Sam
Commented Jun 15 at 10:12
• The fuzzy logic deals with "unsharp" statements like this. Commented Jun 15 at 10:15
• From $P \land \lnot P$, one can conclude anything, let's call it $Q$. No fuzzy logic here, @Peter. From a contradiction (namely \$P \land \lnot P), anything follows, whether true or false. Commented Jun 15 at 17:01
• It seems like you're responding to some existing text or line of argument; can you indicate it? Commented Jun 15 at 19:46