Why this holds? p(X) ⊨ p(f(X))

Question

is there somebody who could help me understand? These statements were given to me to illustrate a logical consequence:

● p(X) ⊨ p(f(X))

● p(X) ⊨ p(f(Y))

Where p is a predicate symbol a f is a function symbol and {X,Y} should be variables. Question is why is that?

Answer 1

There are some slightly different definitions of logical consequence; they all agree on closed formulae (i.e. sentences) but they may disagree on formulae with free variables.

According to the definition in the Wiki's entry referenced, we have that :

$\Gamma \vDash \varphi$

iff :

there is no model $\mathcal{I}$ in which all members of $\Gamma$ are true and $\varphi$ is false. Or, in other words, the set of the interpretations that make all members of $\Gamma$ true is a subset of the set of the interpretations that make $\varphi$ true.

The issue is : what means "to be true" for a formula $\varphi(x)$ with $x$ free ?

According to one approach [see Dirk van Dalen, Logic and Structure (5th ed - 2013), page 67] :

$\varphi$ is true in an interpretation $\mathcal{I}$ iff $CL(\varphi)$ is true in $\mathcal{I}$, where : if $FV(\varphi) = \{ z_1, \ldots , z_k \}$, then $Cl(\varphi) := \forall z_1 \ldots \forall z_k \varphi$ is the universal closure of $\varphi$.

With this definition, a formula $p(x)$ is true in an interpretation $\mathcal{I}$ iff $\forall x p(x)$ is.

Thus, clearly :

$p(x) \vDash p(t)$

for every term $t$ of the language [intuitively, if $p(x)$ holds iff it holds for every object in the domain, then if $p(x)$ holds, then it holds for an object whatever "named by" $t$].

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Example

Let $\mathbb N$ the domain of the interpretation; the predicate $p(x)$ is interpreted as "$x$ is greater or equal to $0$". Let the function $f$ be interpreted as the "successor" function $S$.

Then, in this interpretation, the formula $p(x)$ means : $x \ge 0$.

This formula is true in $\mathbb N$ for every $x$.

With this interpreattion, $p(f(x))$ means : $S(x) \ge 0$.

But $S(x)$ is the successor of $x$, i.e. is $x+1$; thus, the previous formula means : $x+1 \ge 0$, which is again true.

Answer 2

OK I figured out where the problem was.

● p(X) ⊨ p(f(X))

In this form, statement does not hold as Mauro stated. But in my materials there is implicit universal quantificator, which makes it look like:

● (∀X) p(X) ⊨ (∀X) p(f(X))

And from there it is simple, because if every X (member of interpretation domain) has predicate p true, then p(f(X)) must be also true, as f(X) maps to same interpretation domain.

Sorry for incomplete question