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We have the following predicate logical formula $F$: $$\forall x(E(x,y)\rightarrow \neg \exists z(E(f(x,z),y)\land E(y,z)))$$

I want to give a substitution $\sigma$ that is not collision free for $F$.

Could you give me some hints how we could do that?

$$$$

EDIT:

Then I want to do the following:

Give an interpretation $(D_1,I_1)$ and a variable assignment $\beta_1$ so that it holds that $\text{val}_{D_1, I_1,\beta_1}(F) =w$.

By the definition we have that $\text{val}_{D, I,\beta}(1)=w$ and $\text{val}_{D, I,\beta}(0)=f$.

We want that $\text{val}_{D_1, I_1,\beta_1}(F) =w$.

We have that $\text{val}_{D_1, I_1,\beta_1}(x)=\beta_1 (x)$, right?

Does this have to be equal to $w$ ? Does this mean that it must hold that $\beta_1(x)=1$ ?

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    The leading quantifier $\forall x$ binds all occurrences of $x$ in the formula. The existential quantifier $\exists z$ binds all occurrences of $z$ in the formula. The three occurences of $y$ are *free*: thus, we may subst for $y$. – 2017-01-18
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    If we hav no rules for building "complex" terms, we have to use variables; thus, we have to replace $y$ with a new var : $u$ and the subst will be "collision free". – 2017-01-18
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    For a subst that is **not** collision free", a very simple one is : $x$ in place of $y$. The result will be : $∀x(E(x,x)→¬∃z(E(f(x,z),x)∧E(x,z)))$ and we have that the three "new" occurrences of $x$ have been captured by the leading quantifier $∀x$. – 2017-01-18
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    So, the substitution $y=x$ is not collision free because now there is a quantifier $\forall x$ and $y$ is not free anymore. Is this correct? @MauroALLEGRANZA – 2017-01-18
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    Not clear; let $\phi$ the formula; the subst $\sigma(y)=x$ (i.e. put $x$ in place of $y$) will produce the formula $\phi[x/y]$ that is : $∀x(E(x,x) \to \ldots$ where the new occurrences of $x$ have been captured. – 2017-01-18
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    I see!! Then I want to do the following: Give an interpretation $(D_1, I_1, )$ and a variable assignment $\beta_1$ so that it holds that $\text{val}_{D_1, I_1,\beta_1}(F) =w$. By the definition we have that $\text{val}_{D, I,\beta}(1)=w$ and $\text{val}_{D, I,\beta}(0)=f$. We want that $\text{val}_{D_1, I_1,\beta_1}(F) =w$. We have that $\text{val}_{D_1, I_1,\beta_1}(x)=\beta_1 (x)$. Does this have to be equal to $w$ ? Does this mean that it must hold that $\beta_1(x)=0$ ? @MauroALLEGRANZA – 2017-01-18
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    Do you have an idea @MauroALLEGRANZA ? – 2017-01-18
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    What are $w$ and $f$ ? What are you trying to do ? – 2017-01-18
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    w=True, f=False. So, we have to find a variable assignment so that the formula is true or false? But how can we do that ? @MauroALLEGRANZA – 2017-01-18
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    But this has nothing to do with collision-free and substitution of a term for a variable into a formula... – 2017-01-18
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    These two are in the same question, so I thought maybe they are related to each other. Do you maybe have an idea how we could find the variable assignment? @MauroALLEGRANZA – 2017-01-18
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    Is the domain $\{0,1\}$ ? @MauroALLEGRANZA – 2017-01-18

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You have to find an interpretation (domain $D$ plus an interpreting $I$ for the binary predicate symbol $E$ and the binary function symbol $f$) that satisfy the formula $F$.

Due to the free occurrences of the var $y$ in $F$ you have to use a variable assignment function $β$ (that assigns an element of the domain $D$ to $y$) such that $F$ is true in $D$ with $I$ and $β$, i.e.: $D,I,β \vDash F$.

An "easy trick" is to falsify the antecedent of the conditional.

We can consider $D = \{ 0,1 \}$ as domain and interpret $E$ as: $>$ ("greater than").

Let $\beta(y)=0$; thus $E(x,y)$ is interpreted with $\beta$ as $x > 0$, which is not true for all numbers in $D$.

Thus, using $\beta$, the formula is interpeted as:

$\forall x [(x > 0) \to \lnot \exists z (f(x,z) > 0) \land (0 > z))]$

that is true in $D$, irrespective of the interpretation of $f$; i.e.:

$D, I, \beta \vDash F$.