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I want to interpret the below sentences...

  1. ∀x∃y(Cube(x) → (Tet(y) ∧ LeftOf(x, y)))
  2. ∃y∀x (Cube(x) → (Tet(y) ∧ LeftOf(x, y)))

Actually I could interpret the first sentence.

Meaning: Every Cube is left of some Tet

But.. I can't interpret the second sentence.

Let (Cube(x) → (Tet(y) ∧ LeftOf(x, y))) be P(x, y).

There exits some y such that for all x P(x, y) is ture?

.. I don't know.......

please help me to know about it!!

Jonghwan Hyeon
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2 Answers2

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Take it one step at a time. You have $$\exists y\forall x\left(\operatorname{Cube}(x)\to\Big(\operatorname{Tet}(y)\land\operatorname{LeftOf}(x,y)\Big)\right)\;.\tag{1}$$ The first step in translating it is straightforward: just expand the quantifiers and predicates into words.

(a) There is a $y$ such that for every $x$, if $x$ is a cube, then $y$ is a tetrahedron and $x$ is to the left of $y$.

Now notice that if there is at least one cube, the statement y is a tetrahedron does not depend in any way on $x$. We could just as well say:

(b) There is a $y$ such that $y$ is a tetrahedron, and for every $x$, if $x$ is a cube, then $x$ is to the left of $y$.

It would mean the same thing.

The qualifier if there is at least one cube is necessary, because if there are no cubes, $(1)$ and (a) are vacuously true. To keep matters simple, I’ll assume for the moment that there is at least one cube and derive a simple English equivalent; once we have that, we can modify it to get rid of the extra assumption.

Now we can start turning (b) into more ordinary English:

(c) There is a tetrahedron such that every cube is to the left of that tetrahedron.

Or in even more straightforward English:

(d) There is a tetrahedron that is to the right of every cube.

Remember, though, this was equivalent to $(1)$ and (a) only on the extra assumption that there is at least one cube, and that $(1)$ and (a) are vacuously true if there are no cubes. To make (d) equivalent to $(a)$, we must add an alternative:

(e) Either there are no cubes, or there is a tetrahedron that is to the right of every cube.

Brian M. Scott
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  • Wow.. it's really detail :D. Thank you! – Jonghwan Hyeon May 15 '12 at 18:21
  • This is not quite right. Your (a) and (b) are only equivalent under the additional assumption that there is a cube, that is, $\exists x, \text{Cube}(x)$. If there are no cubes, then there is no requirement that $y$ be a tetrahedron, since the implication is vacuously satisfied. – Alex Kruckman May 15 '12 at 19:39
  • Hmm.. sorry.. but I don't know the difference between first and second sentence.. I think the both have same meaning, although their english sentences are different.. is that right? – Jonghwan Hyeon May 15 '12 at 23:12
  • @JonghwanHyeon: Not quite. Suppose that there are no cubes and no tetrahedrons; then (a) is vacuously true, and (b) is false. – Brian M. Scott May 16 '12 at 03:20
  • @Brian M. Scott: Oh I mean, between ∀x∃y(Cube(x) → (Tet(y) ∧ LeftOf(x, y))), and ∃y∀x (Cube(x) → (Tet(y) ∧LeftOf(x, y))) – Jonghwan Hyeon May 16 '12 at 03:23
  • @JonghwanHyeon: Okay; that’s covered by your new question, which now has a couple of good answers. – Brian M. Scott May 16 '12 at 03:25
  • @BrianM.Scott Your adjustment still claims that (a) is equivalent to (d). This isn't correct. If there are no cubes, (a) simply asserts that there is an element $y$ in the universe, but (d) asserts that there is an element $y$ in the universe which is a tetrahedron. Sorry to be nitpicky, but I think it's important to be careful about the edge cases in basic logic like this. – Alex Kruckman May 16 '12 at 03:26
  • @Alex: You’re right: I simply lost track of which one of (a) and (b) actually matched the formalism. I’ll take care of it in a little bit. – Brian M. Scott May 16 '12 at 03:27
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The exact paraphrase of the second sentence would be: there exists y such that for all x if x is a cube than y has property tet and it holds that leftof(x,y). Also your abbreviation seems correct.

borg
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