The Conditional and Biconditional Connectives

If you only learn one connective deeply, make it the conditional. Almost every mathematical theorem has the shape "if $P$, then $Q$" — from elementary geometry ("if a triangle has three equal sides, then it has three equal angles") to advanced analysis ("if a sequence is bounded and monotone, then it converges"). The conditional $P \Rightarrow Q$ has a surprisingly subtle truth table that catches almost every beginner off guard, and the related notions of converse, inverse, and contrapositive are essential vocabulary for the rest of the book. Master them now and the proof techniques of Chapter 3 will feel natural; skip them and proofs will feel like a foreign language.

The truth table of $P \Rightarrow Q$

The conditional $P \Rightarrow Q$ is read "if $P$, then $Q$" or "$P$ implies $Q$." It is false only in one row of its truth table — when $P$ is true but $Q$ is false — and true in every other row.

The two F-rows of $P$ look strange: how can $P \Rightarrow Q$ be true when $P$ is false? The answer is vacuous truth: a conditional with a false antecedent makes no claim about reality, and by default a claim that says nothing is counted as true. "If 2 + 2 = 5, then I am the King of France" is technically true.

Converse, inverse, contrapositive

Three related conditionals are built from $P \Rightarrow Q$ by swapping or negating its parts:

Converse: $Q \Rightarrow P$ (swap antecedent and consequent).

Inverse: $\neg P \Rightarrow \neg Q$ (negate both, keep the order).

Contrapositive: $\neg Q \Rightarrow \neg P$ (swap and negate — both moves at once).

Critical fact: the contrapositive is logically equivalent to the original, while the converse is not. You can verify this in two seconds by comparing the truth-table columns of $P \Rightarrow Q$ and $\neg Q \Rightarrow \neg P$ — they match in every row. The contrapositive equivalence is the backbone of proof by contrapositive, one of the workhorse proof techniques in Chapter 3.

The biconditional

The biconditional $P \Leftrightarrow Q$ is read "$P$ if and only if $Q$" (often abbreviated iff). It is true exactly when $P$ and $Q$ have the same truth value — both true or both false. Logically,

Proving a biconditional is therefore a two-part task: show $P \Rightarrow Q$ and show $Q \Rightarrow P$.

The language of conditions

English contains many ways of saying the same conditional, and confusing them is a major source of student errors:

"$P$ is sufficient for $Q$" means $P \Rightarrow Q$ (knowing $P$ is enough to conclude $Q$).

"$P$ is necessary for $Q$" means $Q \Rightarrow P$ (without $P$, $Q$ cannot happen).

"$P$ only if $Q$" means $P \Rightarrow Q$ (the only way to have $P$ is to also have $Q$).

"$P$ if and only if $Q$" means $P \Leftrightarrow Q$.

The two phrases that trip people up most are "only if" (which puts $Q$ after the arrow, not before) and "necessary" (which reverses the arrow). Slowing down on these phrases pays for itself for the rest of the course.

Pause and think: The statement "If the moon is made of cheese, then $2 + 2 = 4$" is true (vacuously, because the antecedent is false — and incidentally the consequent happens to be true). Now consider "If $2 + 2 = 4$, then the moon is made of cheese." Is this also true? (Hint: check which row of the truth table this corresponds to.)

Try it

• Before writing: predict the contrapositive of "If $x > 3$, then $x^2 > 9$." Then check whether your contrapositive is equivalent to the original by finding a value of $x$ that makes the original true.
• Pick the converse, inverse, and contrapositive of "If a quadrilateral is a square, then it is a rectangle." Decide which of the three is necessarily true and which is not.
• Open the truth-table widget. Build the columns of $P \Rightarrow Q$ and $\neg Q \Rightarrow \neg P$. Compare row by row.
• Translate to symbolic form: "Differentiability is sufficient for continuity." Now translate "Continuity is necessary for differentiability." Are the two translations equivalent?
• Predict first: is "$n$ is even if and only if $n^2$ is even" a biconditional that is true for all integers $n$? Test a few values, then think about both directions.

A trap to watch for

The cardinal sin of conditional logic is confusing a statement with its converse. "If you live in Paris, you live in France" is true. "If you live in France, you live in Paris" is false — the converse fails because not every French resident lives in the capital. A theorem and its converse are different theorems; proving one tells you nothing about the other. Whenever a problem asks for a biconditional or an "iff," you must prove BOTH directions — $P \Rightarrow Q$ AND $Q \Rightarrow P$ — or you have only done half the work.

What you now know

You can read, write, and manipulate conditionals and biconditionals; you can produce converse, inverse, and contrapositive on demand; and you have the vocabulary "necessary," "sufficient," and "only if" under control. Chapter 1 is now complete — you have a full toolkit for propositional logic. Chapter 2 lifts that toolkit to predicate logic: statements about every $x$ and statements about some $x$. That is where mathematics really takes off.

References

• Velleman, D. J. (2019). How to Prove It: A Structured Approach (3rd ed.). Cambridge University Press, ch. 1.
• Enderton, H. B. (2001). A Mathematical Introduction to Logic (2nd ed.). Academic Press, §1.2.
• Smullyan, R. M. (1995). First-Order Logic. Dover Publications, ch. 1.
• Mendelson, E. (2015). Introduction to Mathematical Logic (6th ed.). Chapman and Hall/CRC, ch. 1.