Verifying logical equivalence in first order logic

Question

I am given this definition:

Let the domain of discourse $D$ be the set of all people and houses, the unary relation symbol $P$ is the set of people, and the unary relation $H$ is the set of houses, and and the binary relation $A$ is always interpreted as the set of pairs $(p,h)$ such that $p$ is a person,$h$ is a house and person $p$ has access to house $h$.

Question:

Express the following using an appropriate sentence of first order logic:

No house can be accessed by more than one person.

Attempt:

I attempted expressing the sentence and came up with this:

$\forall x(H(X) \Rightarrow \forall y\forall z ((P(y) \land P(x) \land A(y,x) \land A(z,x))\Rightarrow z=y))$

While the answer given is this:

$\forall x \forall y ((P(X) \land P(y) \land x \neq y)\Rightarrow \neg \exists z(H(z) \land A(x,z) \land A(y,z)))$

Is there any way of deducing whether or not the two sentences are describing the same thing?

Answer 1

In order to "compare" the two formulae, we have to re-write yours as follows :

$∀z[H(z) \to ∀x∀y((P(y)∧P(x)∧A(x,z)∧A(y,z)) \to x=y)]$

that is equivalent to :

$∀z∀x∀y[H(z) \to (P(y)∧P(x)∧A(x,z)∧A(y,z)) \to x=y)]$

becuase $x,y$ are not free in $H(z)$.

The answer is :

$∀x∀y[(P(x)∧P(y)∧x≠y) \to ∀z¬(H(z)∧A(x,z)∧A(y,z))]$

i.e.

$∀x∀y∀z[(P(x)∧P(y)∧x≠y) \to ¬(H(z)∧A(x,z)∧A(y,z))]$.

We have now to "compare" the two "propositional" parts.

Yours is :

$\lnot H(z) \lor \lnot (P(y)∧P(x)∧A(x,z)∧A(y,z)) \lor x=y$

i.e.

$\lnot H(z) \lor \lnot P(y) \lor \lnot P(x) \lor \lnot A(x,z) \lor \lnot A(y,z) \lor x=y$.

The answer is :

$\lnot (P(x)∧P(y)∧x≠y) \lor ¬(H(z)∧A(x,z)∧A(y,z))$

i.e.

$\lnot P(x) \lor \lnot P(y) \lor x=y \lor \lnot H(z) \lor \lnot A(x,z) \lor \lnot A(y,z)$.

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In conclusion : YES, they are the same.