Predicate logic example

Question

While studying predicate logic, i see some example as exercise there.But can't figure it out.Can anyone help me?
(i)If a brick is on another brick, it is not on the table.
(ii)Every brick is on the table or on another brick.
(iii) No brick is on a brick which is also on a brick.

EDIT:
My answers are:
i) ∀x∃y brick_on(x,y) -> ¬brick_on(x,table)
ii) ∀x brick(x) -> onTable(x) ∨ (∃y on(x,y))
iii) ∀x∃y∃z brick_on(x,y) -> ¬brick_on(z,x)

Could anyone tell me these are right or wrong?

Answer 1

In your posted work, you've got a great start, in terms of the logic you used. However, we need to "clean up" parentheses, make a few corrections, define the notation you are using, and use this notation consistently

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Let's start with defining notation, and then let's stick with it.

To simplify matters, we'll let the domain of our universe consist of bricks.

Let $T(x)$ denote "x is on the table".

Let $O(x, y)$ denote "x is on top of y".

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Now to the translations: Pay particular attention to the parentheses, which are used to enclose everything which is within the scope of a quantifier which precedes the parentheses.

First sentence: (i) If a brick is on another brick, it is not on the table.

$$(i)\quad \forall x \Big(\exists y \big(O(x, y)\big) \rightarrow \lnot T(x)\Big)$$

Then we have (ii) Every brick is on the table or on another brick:

$$(ii)\quad \forall x\,\Big(T(x) \lor \exists y\,\big(O(x, y)\big)\Big)$$

So far, your work has been very close to the above. Now let's look at the final sentence:

(iii) No brick is on a brick which is also on a brick.

In loglish: "For all bricks x, if there exists a brick y such that O(x, y), then there does not exist any z such that O(y,z)." Now, the full translation:

$$(iii)\quad \forall x \Big(\exists y \big(O(x,y) \rightarrow \lnot\exists z(O(y,z)\big)\Big)$$

Or alternatively, we can say in loglish "For all bricks x, if there exists a brick y such that O(x, y), then for all bricks z, it is not the case that O(z,x)":

$$(iii)\quad \forall x \Big(\exists y\, \big(O(x,y) \rightarrow \forall z(\lnot O(y,z)\big)\Big)$$

Answer 2

Let $B$ be the set of bricks, and define $T(b)$ to be the predicate declaring $b$ is on a table, and $S(b,b')$ to be that declaring the brick $b$ is on top of the brick $b'$. Now rephrase the statements in terms of these new things, firstly in plain English in your head, then in logical notation.