Translating If, Only If, and Unless into Implications

Symbolizing English conditionals is a recurring konkur skill, and almost every trap comes from one phrase: only if. The reliable approach is to memorize which English connective marks the sufficient condition (the hypothesis, antecedent) and which marks the necessary condition (the consequent).

The three core patterns

Let $p$ and $q$ be the two simple statements.

Notice that if and only if point the arrow in opposite directions. "$p$ if $q$" puts $q$ on the left; "$p$ only if $q$" puts $q$ on the right. Combining both gives the biconditional: "$p$ if and only if $q$" is $p \leftrightarrow q$.

Why "only if" is $p \to q$

"$A$ only if $B$" asserts that the only way $A$ can be true is for $B$ to be true — so $B$ is a necessary condition for $A$. Necessary conditions always sit in the consequent: $A \to B$. Equivalently, if $B$ fails then $A$ must fail, i.e. $\neg B \to \neg A$ (the contrapositive), which is the same proposition.

Why "unless" is $\neg q \to p$

Read "unless" as "if not": "$p$ unless $q$" $=$ "$p$ if not $q$" $= \neg q \to p$. By the contrapositive (or by $\neg q \to p \equiv q \vee p$) this also equals $\neg p \to q$ — handy when the options are written the other way around.

How to apply (3 steps)

1. Name the atoms and find the connective (if / only if / unless / if and only if).
2. Place the arrow using the table. Compound parts like "solve all exercises and stay focused" become a single side, e.g. $q \wedge r$.
3. Match against options up to equivalence. If the literal form is absent, convert via the contrapositive $X \to Y \equiv \neg Y \to \neg X$ or the disjunction $X \to Y \equiv \neg X \vee Y$, then push negations inward with De Morgan: $\neg(q \wedge r) \equiv \neg q \vee \neg r$.

Mini-example

Symbolize "You pass only if you study and sleep," with $p$ = pass, $q$ = study, $r$ = sleep.

• "only if" $\Rightarrow$ the part after it is necessary $\Rightarrow$ consequent: $p \to (q \wedge r).$
• If no option shows this directly, take the contrapositive: $\neg(q \wedge r) \to \neg p,$ and (if needed) apply De Morgan to get $(\neg q \vee \neg r) \to \neg p$. All three are the same proposition.

Exam trick: if and only if are not interchangeable — "$p$ if $q$" is $q \to p$ but "$p$ only if $q$" is $p \to q$. When your derived $p \to (q \wedge r)$ is missing from the choices, don't panic: its contrapositive $\neg(q \wedge r) \to \neg p$ (often shown after De Morgan as $(\neg q \vee \neg r)\to\neg p$) is logically identical and is usually the intended answer.