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Basically I've this program, in which I can create shapes like pentagon, triangle, and square. I need, to create a geometric world such that these two following statements are true:

$ ¬(∀x)(∀y)(∀z)\{ (∃t)[BackOf(x,t) ∧ BackOf(y,t) ∧ BackOf(z,t) ∧ Square(t)] → x = y ∨ x = z ∨ y = z \}; $

$(∀x)(∀y)(∀z)\{ [BackOf(x,a) ∧ BackOf(y,a) ∧ BackOf(z,a) ∧ Square(a)] → x = y ∨ x = z ∨ y = z \} $

Now a bit of an explanation: x,y,z are variables that represent shapes in the world, while the letter a is a constant, meaning there exists a shape named a (not necessary a square - but must be a shape named a).

BackOf(x,y) is a function that returns true if x is a shape that is found behind y.

Square(x) is a function that returns true if x is a square. Now all the other predicates are well known to you guys and also the symbols of "or - V" and "derive ->".

Basically I construct this world by creating shapes and giving them the needed names (like a), I draw triangles, pentagons and squares on a table.

Now hopefully that you understood the program and my task: I'm having a difficulty creating a geometric world in which these 2 sentences receive true value since I'm almost positive they are the negation of each other.

Basically, the second statement will be able to receive false value only if the part before the derive sign:

$(∀x)(∀y)(∀z)[BackOf(x,a) ∧ BackOf(y,a) ∧ BackOf(z,a) ∧ Square(a)]$

is true, and that only happens when a itself is behind a because of the predicates "for all" before the statement and that is impossible for a shape to be behind itself, therefore the second statement is always true because false ->true/false is true (I'm pretty sure of it).

The first statement is something I've not yet understand completely and I could use some help to distinguish it.

Thanks!

1 Answers 1

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The first statement says that it is not true that there cannot be three or more different objects behind any square.

The second statement says that if $a$ is a square, then there cannot be three or more different objects behind it.

OK, to satisfy this, simply make sure that $a$ is not a square, and have another object in the world that is a square, and that has at least three different object behind it.

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    First of all, you are brilliant, since it solved it and I've done it. Now, I still don't fully understand it, could you elaborate? – 2017-01-06
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    @BarVered The second statement is a universal statement with a conditional inside of it, rather than a conditional whose 'if' part is a universal. So, your analysis of this sentence was incorrect. – 2017-01-06
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    @BarVered The second statement says: 'If you ever have three objects behind square $a$, then at least two of those three are the same' ... Which really means: if $a$ is a square, then it cannot have three different objects behind it. But the first sentence says that 'it is not true that (if there is a square with three objects behind it, then at least two of them are the same)' ... Which means that there is a square with three distinct objects behind it. ... But this square cannot be $a$ according to the second sentence. So: either $a$ is not a square, or $a$ does not have 3 objects behind it – 2017-01-06
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    So you can try this too: have 4 objects $a$, $b$, $c$, and $d$, have them all be squares, but make sure that $a$ is not in front of all the others. – 2017-01-06
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    ... But still have one square in front of all the others, e.g. Have square $d$ in front of squares $a$, $b$, and $c$. – 2017-01-06