Predicate-Logic Translation and Quantifier Rules

Translating English (or Persian) into first-order logic is mechanical once you know two pattern rules and two algebraic laws. Most exam errors come from mixing up the restriction pattern or reversing an implication.

1. The two restriction patterns

A restricted quantifier ("all $A$ are $B$", "some $A$ is $B$") always pairs a quantifier with a fixed connective:

The rule: $\forall$ goes with $\to$, and $\exists$ goes with $\wedge$. Using $\exists x\,(A(x) \to \dots)$ is almost always wrong, because $A(x) \to \dots$ is vacuously true for any $x$ outside $A$, so the formula no longer says "some $A$".

So "there is a factory that sells cars only to plateless people" begins $\exists x\,[K(x) \wedge \dots]$, not $\forall x\,[K(x) \to \dots]$ — the latter would assert the property of every factory.

2. Quantifier distribution over connectives

Quantifiers distribute over only one connective each, and the match is the natural pairing $\forall/\wedge$ and $\exists/\vee$:

$\forall x\,(P(x) \wedge Q(x)) \;\equiv\; \forall x\,P(x) \wedge \forall x\,Q(x),$

$\exists x\,(P(x) \vee Q(x)) \;\equiv\; \exists x\,P(x) \vee \exists x\,Q(x).$

The mismatched pairs give only a one-way implication, never equivalence:

$\exists x\,(P(x) \wedge Q(x)) \;\Rightarrow\; \exists x\,P(x) \wedge \exists x\,Q(x) \quad(\text{converse fails}),$

$\forall x\,P(x) \vee \forall x\,Q(x) \;\Rightarrow\; \forall x\,(P(x) \vee Q(x)) \quad(\text{converse fails}).$

The failure always has the same cause: the two separate quantifiers may pick different witnesses. For $\exists$ over $\wedge$, take domain $\{1,2\}$ with $P(1),\,\neg P(2),\,\neg Q(1),\,Q(2)$. Then $\exists x\,P(x)$ and $\exists x\,Q(x)$ are both true (witnesses $1$ and $2$), yet no single $x$ satisfies $P(x) \wedge Q(x)$, so $\exists x\,(P(x) \wedge Q(x))$ is false — the formulas are not equivalent.

3. Getting the implication direction right

Many trap options keep the right shape but reverse an implication. Two keywords to watch:

• "only": "$x$ does $P$ only to $Q$s" means doing $P$ forces being a $Q$, i.e. $P \to Q$ — not $Q \to P$. So "sells cars only to plateless people" is $\big(C(y) \wedge S(x,y,z)\big) \to \big(P(z) \wedge \neg H(z)\big)$, with the act of selling as antecedent. Writing $\big(P(z) \wedge \neg H(z)\big) \to \big(C(y) \wedge S(x,y,z)\big)$ instead says "every plateless person is sold a car" — a different claim.
• "unless": "$P$ unless $Q$" $\equiv P \vee Q \equiv \neg Q \to P$. The condition is the negated clause, so the implication runs from $\neg Q$, never from $Q$.

Mini-example

Translate "every student who studies passes": the subject is restricted ($\forall$ with $\to$), and "who studies" is an extra restriction folded into the antecedent:

$\forall x\,\big(\big(\mathrm{Student}(x) \wedge \mathrm{Studies}(x)\big) \to \mathrm{Passes}(x)\big).$

Note both restrictions sit on the left of the $\to$, joined by $\wedge$ — the universal pattern.

Exam trick: Lock in the four pairings: $\forall$ with $\to$, $\exists$ with $\wedge$ for restriction; $\forall$ over $\wedge$, $\exists$ over $\vee$ for distribution. Any other combination ($\exists$ with $\to$, $\forall$ with $\wedge$ as a restriction, or distributing $\exists$ over $\wedge$ / $\forall$ over $\vee$ as an equivalence) is a planted wrong answer.