# How to correctly negate a predicate bounded by some quantifiers?

this is a problem which was asked in GATE CS 2010.

This is question statement:
Q:
Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x, y, t))?

Options:
A: Everyone can fool some person at some time.
B: No one can fool everyone all the time.
C: Everyone cannot fool some person all the time.
D: No one can fool some person at some time.

According to my solution:
If F(x): person x can fool person y at time t.
Then
$$\forall$$x $$\exists$$y $$\exists$$t ( ¬F( x, y, t ) )
is same as "Not all person x can fool some person y at some time t. which can be rewritten as "No one can fool some person at some time".
Hence Option D must be the correct one.
However I am wrong.

How to approach these type of problems.

$$\forall x\ \exists y\ \exists t\ [\neg F(x,y,t)]\equiv \forall x\ \exists y\ [\neg\ \forall t F(x,y,t)]\equiv \forall x\ [\neg\ \forall y\ \forall t F(x,y,t)]\equiv \neg [\exists x\ \forall y\ \forall t F(x,y,t)]$$
• Yes, so long as you correctly invert the operators encountered (e.g. $\neg\ \exists x$ becomes $\forall x\ \neg$), and distribute when appropriate (e.g. $\neg (A \lor B)$ is $(\neg A) \land (\neg B)$) – eru-cs Oct 7 '19 at 10:29