First post, apologies if it's in the wrong format etc. I have this question from my lecturer and am really confused on how I would go about solving this. If anyone can provide any assistance or examples that would great.
Definitions of predicates:
V (x): x is a point (or vector).
L(x): x is a line.
X(x, y, z): the lines x and y intersect at point z.
P(x, z): z is a point on line x.
A(x, y, z): adding vector y to line x results in line z.
Based on these and also using the equality predicate = on vectors and lines; ~0 is the 0 vector. Translate the following into English.
- ∀x. L(x) → ∃y. P(x, y)
- ∀a. ∀b. ∀c. ∀d. (X(a, b, c) ∧ X(a, b, d)) → c = d ∨ a = b
- ∀x. L(x) → ((∃y. V (y) ∧ P(x, y) ∧ A(x, y, x)) ↔ P(x,~0))
Thank you for any help you can provide!
Take the first expression (which, by the way, is missing parentheses):
$$\forall x (L(x) \rightarrow P(x,y))$$
The $\forall x$ indicates that $L(x) \rightarrow P(x,y)$ is true for all $x$, i.e. for every object. So, it says
$$\text{"For every object it is true that if it is a line, then $\exists y P(x,y)$"}$$
OK, so what does that latter part say? Here, the $\exists y$ indicates that there is some object $y$ for which $P(x,y)$ is true, i.e. for which $y$ is a point on $x$. So, filling this in, we get:
$$\text{"For every object $x$ it is true that if it is a line, then there is some object $y$ that is a point on $x$"}$$
Now that is a very unnatural statement, but you can paraphrase it like this:
$$\text{"For every line $x$, there is some object that is a point on $x$"}$$
And that can be further simplified to:
$$\text{"Every line has a point on it"}$$
You also mention:
$$\exists x \exists y P(x,y)$$
This translates to:
$$\text{"There is some object $x$ such that there is some object $y$ such that $y$ is a point on $x$}$$
... which rephrased would indeed be
$$\text{"Some line(s) have some point(s) on them"}$$
(The (s) is to emphasize that $\exists y$ is looking for 'one or more')