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Consider the well-formed formual below, What do you get by substituting t for x?

$\forall x (p(x,y) \land \forall y(p(y,x) \rightarrow \exists x (p(x,z) \land q(z)))$

are there any restrictions or is it just a straight substitute yielding

$\forall t (p(t,y) \land \forall y(p(y,t) \rightarrow \exists t (p(t,z) \land q(z)))$

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    The subst replace *free* occurrences of $x$ with $t$ and **not** the bound ones. Thus, you cannot replace the occurrence of $x$ in $\forall x$ and $\exists x$ and neither the occurrences of $x$ in the *scope* of a quantifier $\forall x$ and $\exists x$. – 2017-02-23

1 Answers 1

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Def of substitution :

For atomic $α$, $α(t/x)$ is the expression obtained from $α$ by replacing the variable $x$ by $t$.

$(¬α)(t/x)$ is $¬α(t/x)$

...

$(∀y \ α) (t/x)$ is $∀y \ α$, if $x=y$, and $∀y \ α(t/x)$, if $x \ne y$ (and the same for $\exists$).

Thus, in order to know:

What do you get by substituting $t$ for $x$ into : $∀x(p(x,y)∧∀y(p(y,x)→∃x(p(x,z)∧q(z))))$

we have to apply the above definition (the clause for $\forall$) and we have that:

$(∀x(p(x,y)∧∀y(p(y,x)→∃x(p(x,z)∧q(z)))))(t/x)$

is:

$∀x(p(x,y)∧∀y(p(y,x)→∃x(p(x,z)∧q(z))))$.