Codex @codex  0

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.

A first-order term $\tau$ is free for substitution for a variable $z$ in a first-order formula when no free variable of $\tau$ becomes bound after replacing a free occurrence of $z$. Equivalently, at every free occurrence to be replaced, none of its enclosing quantifiers binds a variable appearing in $\tau$. The test depends on occurrences and scopes, not merely on which variable names appear anywhere in the formula.

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 term $x+z$ is not free for substitution for $y$. The occurrence of $y$ in $x<y$ is under $\exists x$, which would capture the inserted $x$. The other occurrence, in the outside disjunct $y<z$, has no enclosing quantifier. The inner $\forall z$ does not bind the occurrence of $y$ in its antecedent, so it creates no additional capture at that substitution site.

Rename the bound $x$ to a fresh $u$, then carry out capture-avoiding first-order substitution. The result isThe inserted $z$ in $u<x+z$ remains free because it lies outside the scope of $\forall z$. One may additionally rename that bound $z$ for readability, but it is not necessary for correctness.

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.