Converse Inverse and Contrapositive

Changing a Conditional

In Implication and Conditional Statements, we established that \(p\to q\) is false exactly when \(p\) is true and \(q\) is false. We also saw that “If \(x>2\), then \(x>0\)” holds for every real \(x\), while reversing those two inequalities gives a false general claim.

Reversing a conditional is one of three standard transformations. The other two involve negating its components. Although all three produce related statements, only one is always logically equivalent to the original. Knowing which transformation preserves meaning is essential both when reading a theorem and when deciding how to prove it.

We continue to use \(p,q,r\) for proposition letters and \(A,B\) for formulas, including compound formulas. The symbol \(\equiv\) denotes logical equivalence: agreement under every truth assignment.

Three Related Statements

Definitions. For a conditional \(p\to q\), its converse is \(q\to p\), its inverse is \(\neg p\to\neg q\), and its contrapositive is \(\neg q\to\neg p\).

Statement	Form	Change from the original
Original	\(p\to q\)	None
Converse	\(q\to p\)	Exchange hypothesis and conclusion.
Inverse	\(\neg p\to\neg q\)	Negate both; keep their order.
Contrapositive	\(\neg q\to\neg p\)	Exchange them and negate both.

The names are relative to the chosen original statement. They describe how a new conditional is formed, not whether it is true. These definitions apply equally to \(A\to B\): the contrapositive is \(\neg B\to\neg A\), with each negation applying to an entire component.

Worked Example: Keeping the Boundaries

For a real number \(x\), take the original conditional to be “If \(x>2\), then \(x>0\).” Its related forms are:

Statement	In words
Converse	If \(x>0\), then \(x>2\).
Inverse	If \(x\leq2\), then \(x\leq0\).
Contrapositive	If \(x\leq0\), then \(x\leq2\).

The negation of \(x>2\) is \(x\leq2\), not \(x<2\). Similarly, the negation of \(x>0\) is \(x\leq0\). Equality must remain in each negated inequality.

The original holds for every real input by the usual order relation. Its contrapositive does too: any \(x\leq0\) also satisfies \(x\leq2\). But \(x=1\) disproves both the converse and the inverse. For the converse, \(1>0\) is true and \(1>2\) is false. For the inverse, \(1\leq2\) is true and \(1\leq0\) is false.

Equivalence with the Contrapositive

The agreement between the original and its contrapositive in that example is not a special property of inequalities. It follows from the conditional truth table alone.

Theorem (Contraposition). For any formulas \(A,B\),

(A\to B)\equiv(\neg B\to\neg A).

Proof. Fix any truth assignment to the letters occurring in \(A\) and \(B\). By the definition of a conditional, \(A\to B\) is false exactly when \(A\) is true and \(B\) is false. The conditional \(\neg B\to\neg A\) is false exactly when \(\neg B\) is true and \(\neg A\) is false. Those requirements say precisely that \(B\) is false and \(A\) is true.

Thus the two conditionals are false on exactly the same assignments. On every other assignment both are true. They therefore agree under every assignment, which proves the equivalence.

The rewriting result from the previous tutorial gives another expression of the same relationship:

\begin{aligned} (\neg B\to\neg A) &\equiv \neg\neg B\lor\neg A\\ &\equiv B\lor\neg A\\ &\equiv \neg A\lor B\\ &\equiv (A\to B). \end{aligned}

Here the middle steps use the Double Negation Law and the Commutative Law for disjunction. Each step preserves truth values, as established in Logical Equivalence.

The same restriction, expressed differently. The original says that the hypothesis cannot hold while the conclusion fails. The contrapositive says that if the conclusion fails, the hypothesis must fail. Both exclude exactly the same situation.

The Converse and Inverse Form a Second Pair

Corollary. For any formulas \(A,B\),

(B\to A)\equiv(\neg A\to\neg B).

Thus the converse and inverse of a conditional are logically equivalent to each other.

Proof. Apply the Contraposition Theorem to \(B\to A\). Its contrapositive is \(\neg A\to\neg B\), exactly the inverse of the original \(A\to B\).

The full truth table shows both equivalent pairs. The column labels refer to the original \(p\to q\).

\(p\)	\(q\)	Original	Converse	Inverse	Contrapositive
T	T	T	T	T	T
T	F	F	T	T	F
F	T	T	F	F	T
F	F	T	T	T	T

The original and converse differ on the two rows where \(p\) and \(q\) have different truth values. Consequently, reversing an implication is not a valid general equivalence. Negating both components without reversing them is not a valid general equivalence either.

This does not mean that a true mathematical statement must have a false converse. For example, for every real \(x\), “If \(x>0\), then \(x+1>1\)” and its converse are both true: adding \(1\) proves the first, and subtracting \(1\) proves the second. The point is that the truth of the original alone does not establish the converse; the converse needs its own justification.

Contrapositive Is Not Negation

The inverse and contrapositive both contain negation symbols, but neither is the operation of negating the whole conditional. By the negation result proved in the previous tutorial,

\neg(p\to q)\equiv p\land\neg q.

Operation	Result	Relationship to the original
Take the contrapositive	\(\neg q\to\neg p\)	Always has the same truth value.
Take the inverse	\(\neg p\to\neg q\)	May agree or disagree.
Negate the conditional	\(p\land\neg q\)	Always has the opposite truth value.

Worked Example: Negation Versus Contraposition

At a specified real input \(x\), consider “If \(x>0\), then \(x>2\).” Its contrapositive is “If \(x\leq2\), then \(x\leq0\).” Its negation is “\(x>0\) and \(x\leq2\).”

At \(x=1\), the original and contrapositive are both false, while the negation is true. At \(x=3\), the original is true, the contrapositive is vacuously true, and the negation is false.

If instead we consider the claim that the original holds for every real \(x\), disproving it requires just one input satisfying \(x>0\) and \(x\leq2\). It does not require this conjunction to hold for every input.

Working with Compound Components

When a hypothesis or conclusion contains several connectives, exchange the whole components before simplifying their negations. De Morgan's Laws then determine the correct connective inside each negated expression.

Worked Example: A Conjunctive Hypothesis

Consider \((p\land q)\to r\). Here \(A=p\land q\) and \(B=r\). The three transformations are:

\begin{aligned} \text{Converse:}\quad &r\to(p\land q),\\ \text{Inverse:}\quad &\neg(p\land q)\to\neg r,\\ \text{Contrapositive:}\quad &\neg r\to\neg(p\land q). \end{aligned}

By De Morgan's first law, the contrapositive can be written as

\neg r\to(\neg p\lor\neg q).

It says that if the conclusion fails, at least one part of the original hypothesis fails. It does not say that both parts must fail.

To verify the distinction, take \(p=\mathrm{F}\), \(q=\mathrm{T}\), and \(r=\mathrm{F}\). The original is true because its hypothesis is false. The correct contrapositive is also true. But the incorrect expression \(\neg r\to(\neg p\land\neg q)\) is false: its hypothesis is true and its conclusion is false.

For a disjunctive conclusion, the corresponding transformation is

\bigl(p\to(q\lor r)\bigr) \equiv \bigl((\neg q\land\neg r)\to\neg p\bigr).

Here failure of the whole conclusion \(q\lor r\) means that both \(q\) and \(r\) fail, by De Morgan's second law.

Using Contraposition in a Mathematical Argument

Because a conditional and its contrapositive agree at every input, proving either one throughout a fixed domain establishes the other throughout that same domain. This is the basis of proof by contraposition.

Keep the domain fixed.
Identify the allowed inputs and the whole hypothesis \(A\) and conclusion \(B\).

Form the contrapositive.
Write \(\neg B\to\neg A\), simplifying negations without losing boundary cases.

Establish the new implication.
For an arbitrary allowed input satisfying \(\neg B\), show that \(\neg A\) holds.

Return to the original.
Use the Contraposition Theorem to conclude \(A\to B\) on the same domain.

Worked Example: A Restricted Domain Matters

Claim. For every nonnegative real number \(x\), if \(x^2>4\), then \(x>2\).

Within the domain \(x\geq0\), the contrapositive is: if \(x\leq2\), then \(x^2\leq4\).

Proof. Take any nonnegative real \(x\). Suppose \(x\leq2\). Multiplying this inequality by the nonnegative number \(x\) gives \(x^2\leq2x\). Multiplying \(x\leq2\) by \(2\) gives \(2x\leq4\). Hence

x^2\leq2x\leq4.

This proves the contrapositive when its hypothesis holds. When \(x>2\), that hypothesis is false, so the contrapositive is vacuously true. It therefore holds for every input in the stated domain. By the Contraposition Theorem, the original claim holds there as well.

The domain restriction was retained, not negated. Without it, the claim would be false: \(x=-3\) satisfies \(x^2=9>4\) but not \(x>2\). The contrapositive would also fail at that same input, since \(-3\leq2\) but \(9\leq4\) is false.

Choose an equivalent task, not a different claim. Proving the contrapositive establishes the original conditional. Proving the converse or inverse generally does not. Before beginning an argument, check both the direction of the arrow and the scope of each negation.

Check Your Understanding

Use the conditional truth table, the Contraposition Theorem, and the earlier negation laws to justify your answers.

Write the converse, inverse, and contrapositive of “If \(x>5\), then \(x>1\).” Treat each as a claim about every real \(x\). Which are true, and what counterexample disproves each false claim?
Suppose \(p\to q\) is true and \(q\to p\) is false. Determine the truth values of \(p\), \(q\), the inverse, and the contrapositive of \(p\to q\).
Form the contrapositive of \((p\lor q)\to(r\land p)\). Simplify the negations using De Morgan's Laws, keeping the entire hypothesis and conclusion grouped.
At a specified real \(x\), write both the contrapositive and the negation of “If \(x\geq1\), then \(x>0\).” Evaluate the original, its contrapositive, and its negation at \(x=0\) and \(x=1\).
For every real \(x\), prove “If \(3x+1>7\), then \(x>2\)” by establishing its contrapositive. Include the boundary value \(x=2\) in your reasoning.
A learner claims that because a conditional is true, its inverse must be false. Explain why this is incorrect, and give a mathematical conditional whose original, converse, inverse, and contrapositive all hold throughout the stated domain.