Formally proof the following statement: (any ideas how, its for a research) ¬∀x: X.P ⊢ ∃x: X.P → Q?
Update:
eg... Universal quantification (symbol ∀) - is the notion that something is true for
everything. Example: All assignments are boring will be notated as
∀ a: Assignments ∙boring (a)
∀a is called the quantifier, a is a variable of type Assignments, boring (a) is the
predicate.
That explains ∀x: X.P
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I don't understand your notation. Is X in the set P?
Do you mean ⊢ means models?
Without the negation, it sounds like this means all x in P models the formula if there exists an x in P then Q...
...that doesn't make much sense to me.
The negation means that there exists an X.P such that it does not model the formula.
Edit:
It would really help if you could write down what you are saying in English. Unfortunately the notation is not standard (this was a huge contention of mine when taking such courses). You are actually describing first-order predicate logic, but I really need to know what you are saying to be able to help.
What you are describing doesn't make any sense to me. I can only read this a couple of different ways:
1: http://latex.codecogs.com/gif.latex?%5Cforall%20x%...
In this case, the fact that you only look at x in P then the there exists clause is trivially true and thus Q must be a tautology (always true) or the formula is always false (i.e. Q is sometimes false).
2: The only real way, I can see this making any kind of "sense" is for the following formula: http://latex.codecogs.com/gif.latex?%5Cexists%20x%... or http://latex.codecogs.com/gif.latex?%5Cforall%20x%...
The forall statement and exists statement contradict each other (i.e. it cannot be both or, if it is, then the for all "dominates").
Edit:
What is P (I'm assuming that's the domain of x we should consider). It doesn't make sense to state ∀x in P AND ∃x in P. Here's a sentence explaining this contradiction:
For all students in a class, there exists a student in the class such that this student passed.
There are two ways of interpreting this statement:
1) Assuming we can find a single student that passed, this statement is true. The statement is trivially true because of course for ALL students we can find ANOTHER that passed (the one that passed).
2) Assuming at least one student did NOT pass, then the exists statement says nothing. For all students, there exists another student such that the original passed. If we look at passing students, then we can find others (any other) and this student passed, so the statement is true. However, if we can find a single student that failed, then the statement is false. It's NOT true because we have a failing student combined with any other student which makes the implication false.