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To perform a first-order substitution, first rename each conflicting bound variable to a fresh name, then replace the designated free occurrences. The fresh names must avoid the formula's relevant free names and all names in the inserted first-order term. For instance, substituting $f(u,t)$ for $z$ in $\exists tQ(z,t)$ requires renaming the binder, producing $\exists sQ(f(u,t),s)$ rather than $\exists tQ(f(u,t),t)$. The latter would change the intended meaning by capturing $t$.

Replacing every free occurrence of variable $z$ in a first-order formula by a first-order term $\tau$. Bound occurrences of $z$ are left alone. The operation requires either that $\tau$ be free for substitution at those occurrences or that conflicting bound variables first be renamed, as in capture-avoiding first-order substitution.

The term $x+z$ is free for substitution for $z$. The only free occurrence of the target $z$ is in the outside disjunct $y<z$, beyond both quantifier scopes. The occurrences under $\forall z$ are bound variables and are not replaced. Therefore no renaming is required, andThe new $x$ and $z$ in the last disjunct are free. First-order substitution replaces the original free occurrences once; it does not recursively substitute into the inserted term.

The printed expression has one unmatched closing parenthesis. Remove that final surplus parenthesis and read the first disjunct as $\exists xP(x)$; the other disjunct has outer universal quantification over the displayed implication. To distinguish the two independent bound uses of $x$, rename the first to $u$.

Push the outer negation through the logical disjunction and through the quantifiers. Negating the implication preserves its existential antecedent and negates its consequent. Applying De Morgan's laws and the quantifier negation rules gives the negation normal formEach negation is now directly before an atomic formula. No bound variable has been captured, and no original existential quantification in the antecedent has accidentally been negated twice.

The occurrence of $y$ in $P(x,y)$ lies under the quantifier $\exists x$. Directly inserting the term $x+z$ would capture its free $x$: the result $\exists xP(x,x+z)$ would make the inserted $x$ bound and change the intended dependence on the original free value. Therefore the term is not free for the literal substitution requested.

Perform capture-avoiding first-order substitution by first renaming bound variables to fresh names. Renaming the first binder to $u$ is necessary; renaming the second to $v$ is optional but makes the result especially clear. These renamings give the equivalent formula $(\exists uP(u,y))\vee (\exists vQ(v))$. Substituting only the free occurrences of $y$ now givesBoth $x$ and $z$ in the inserted term remain free variables. The original $\exists z$ has scope only over its own disjunct and would not capture the inserted $z$ in the first disjunct; its optional renaming is not a necessary capture repair. This distinction avoids confusing reuse of a letter elsewhere with binding within the relevant scope.

Use broad scope up to the enclosing parenthesis, consistent with (c). Thus the first conjunct is read as$f(u,t)$ is not free for substitution for $z$. The free occurrence of $z$ in $Q(z,t)$ lies under $\exists t$; inserting $f(u,t)$ there would capture its $t$. Rename that bound variable to a fresh $s$ first, using alpha equivalence. The capture-avoiding first-order substitution is thenEvery occurrence of $z$ in the original formula is free, so all four are replaced. The variable $t$ introduced by the term remains free, and the final $\exists yQ(y,y)$ is unchanged.

Under the same scope convention, $g(t,z)$ is not free for substitution for $y$. The occurrence of $y$ inside $Q(f(x,y),f(z,z))$ is free but lies under $\exists t$, so its replacement would capture the new $t$. Rename this binder to fresh $s$. Substitute only for the free $y$ occurrences in the first conjunct:The two occurrences of $y$ in the final conjunct are bound variables and remain unchanged. The inserted $t,z$ remain free.

If instead one gives the printed $\exists t$ only the scope $Q(z,t)$, this part's free-for-substitution answer becomes yes, because neither free $y$ occurrence then lies under that binder. Under that alternative parsing, the result is explicitlyThe narrower reading does not change the negative answer in (1): its occurrence of $z$ inside $Q(z,t)$ still requires renaming that binder before substitution. The broad-scope answers above give a single consistent parsing across the paper.