From Equivalence to “If–Then”
In Logical Equivalence, we learned to compare formulas under every truth assignment and to replace expressions by equivalent ones. We now apply those tools to the conditional connective introduced earlier in this course. Its importance comes from the structure of mathematical claims: under specified hypotheses, a specified conclusion must hold.
For example, “If \(x>2\), then \(x>0\)” does not assert that \(x>2\). It states a requirement on the conclusion when that hypothesis holds. Keeping the hypothesis separate from the conditional as a whole is essential for reading mathematical arguments.
We continue to use \(p,q,r\) for proposition letters, T and F for truth values, and \(A,B\) for formulas that may themselves contain several connectives.
The Meaning of a Conditional
| \(p\) | \(q\) | \(p\to q\) | Interpretation |
|---|---|---|---|
| T | T | T | The hypothesis holds and the conclusion holds. |
| T | F | F | The hypothesis holds but the conclusion fails. |
| F | T | T | The hypothesis does not hold; the conclusion holds. |
| F | F | T | Neither the hypothesis nor the conclusion holds. |
The conditional rules out just one situation: a true hypothesis together with a false conclusion. It does not require both components to be true. That stronger requirement is expressed by the conjunction \(p\land q\), not by \(p\to q\).
The same definition applies to \(A\to B\). On each assignment, first evaluate the entire formula \(A\) and the entire formula \(B\), and then apply the conditional truth table to their two values.
Why a False Hypothesis Gives a True Conditional
When the hypothesis is false, the conditional is said to be vacuously true. This is ordinary truth under the definition, not an additional truth value. The term identifies why the conditional is true: its hypothesis is not satisfied, so the prohibited true-hypothesis, false-conclusion situation does not occur.
Worked Example: Three Inputs, One Conditional
At each specified real input \(x\), let \(p\) mean “\(x>2\)” and \(q\) mean “\(x>0\).” Evaluate \(p\to q\) at the following inputs.
| \(x\) | \(x>2\) | \(x>0\) | Conditional |
|---|---|---|---|
| \(3\) | T | T | T |
| \(1\) | F | T | T, vacuously |
| \(0\) | F | F | T, vacuously |
At \(x=0\), the conclusion is false, but that does not make the conditional false. Its hypothesis is also false. At \(x=1\), the hypothesis is again false, while the conclusion is true. These two inputs show that a false hypothesis places no restriction on the conclusion's truth value.
At the boundary \(x=2\), the strict inequality \(x>2\) is false and \(x>0\) is true, so the conditional is again vacuously true.
The material conditional also makes no assertion about causation or temporal order. “If \(2+2=4\), then \(7>3\)” is true because both component statements are true; no causal connection is needed. Mathematical implication is determined by truth conditions, not by the suggestion that one event produces another.
Translating Conditional Language
The order of clauses in a sentence is not always the order of the arrow. Identify which statement supplies the hypothesis and which is required when that hypothesis holds.
| Wording | Symbolic form |
|---|---|
| If \(p\), then \(q\). | \(p\to q\) |
| \(q\) if \(p\). | \(p\to q\) |
| \(q\) whenever \(p\). | \(p\to q\) |
| \(p\) only if \(q\). | \(p\to q\) |
| \(p\) is sufficient for \(q\). | \(p\to q\) |
| \(q\) is necessary for \(p\). | \(p\to q\) |
Calling \(p\) a sufficient condition for \(q\) means that \(p\) is enough to guarantee \(q\). Calling \(q\) a necessary condition for \(p\) means that \(p\) cannot hold without \(q\). Both describe \(p\to q\).
In particular, “\(p\) only if \(q\)” excludes \(p\) occurring without \(q\). It therefore excludes \(p=\mathrm{T}\), \(q=\mathrm{F}\), exactly as \(p\to q\) does. By contrast, “\(p\) if \(q\)” has \(q\) as its hypothesis and is written \(q\to p\).
Worked Example: Sufficient Is Not Required
For a real number \(x\), the statement “\(x>0\) if \(x>2\)” has hypothesis \(x>2\) and conclusion \(x>0\):
Thus \(x>2\) is sufficient for \(x>0\), and \(x>0\) is necessary for \(x>2\). But \(x>2\) is not necessary for \(x>0\): the value \(x=1\) is positive without exceeding \(2\). A sufficient condition need not be the only way for the conclusion to hold.
Rewriting an Implication
The previous tutorial established how matching truth-table columns prove equivalence. That method gives a useful expression for the conditional using only negation and inclusive disjunction.
Proof. First compare the formulas on all possible pairs of input truth values.
| \(p\) | \(q\) | \(p\to q\) | \(\neg p\) | \(\neg p\lor q\) |
|---|---|---|---|---|
| T | T | T | F | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
The conditional and the final disjunction agree on all four rows. Now fix any assignment to the letters in \(A\) and \(B\). Their values select one of these four rows, on which the outputs agree. Since the assignment was arbitrary, the equivalence holds for \(A,B\), even if they share letters or contain compound expressions.
This equivalence restates the conditional's requirement: either the hypothesis fails, or the conclusion holds, with both possibilities allowed to occur together.
Worked Example: A Compound Hypothesis
Rewrite \((p\land q)\to r\) without an implication symbol. Apply the theorem to the whole hypothesis \(p\land q\), and then use De Morgan's first law:
The second step uses replacement of equivalent formulas, established in Logical Equivalence. The result is false exactly when \(p\) and \(q\) are both true and \(r\) is false. If either part of the hypothesis fails, the original conditional is vacuously true.
Negating a Conditional
To deny an implication is to assert its failure case, not merely to deny its conclusion. The rewriting theorem makes this precise.
Proof. Replace the implication by its equivalent disjunction, apply De Morgan's second law, and then apply the Double Negation Law:
Each replacement preserves truth values under every assignment, so chaining these equivalences proves the claim.
For example, at a specified real input \(x\), the negation of “If \(x>0\), then \(x>1\)” is
The conjunction retains the hypothesis and denies the conclusion. At \(x=1\), both requirements hold, so the conditional is false. At \(x=0\), the conclusion still fails, but the hypothesis fails as well; this input makes the conditional true and its negation false.
Conditional Claims Across a Domain
So far, evaluating inequalities has meant fixing an input first. A general mathematical claim often says that a conditional holds for every input in a stated domain. Such a claim is false as soon as one allowed input makes its hypothesis true and its conclusion false.
Determine which inputs are allowed, what the hypothesis is, and what the conclusion is.
An input with a false hypothesis cannot disprove the conditional.
Both requirements must hold together to give a counterexample.
Worked Example: A Counterexample and a Non-Counterexample
Consider the claim: “For every real number \(x\), if \(x>0\), then \(x>1\).”
Choose \(x=1/2\). The hypothesis is true because \(1/2>0\), while the conclusion is false because \(1/2\) is not greater than \(1\). This is a counterexample, so the general claim is false.
Choosing \(x=-2\) would not work. Although the conclusion is false there, the hypothesis is also false. The conditional is true at that input. The boundary value \(x=1\), however, is another counterexample: it is positive but does not satisfy the strict inequality \(x>1\).
To establish a conditional for every allowed input, one must instead rule out all such counterexamples. For the claim “For every real \(x\), if \(x>2\), then \(x>0\),” take any real \(x\). If \(x>2\), the usual order relation gives \(x>0\), since \(2>0\). If \(x\) is not greater than \(2\), the conditional is vacuously true. These cases cover every real input.
This is the same distinction emphasized in the previous tutorial. Logical equivalences hold under unrestricted truth assignments; a mathematical conditional may depend on the meaning of its components and the specified domain. Neither a few successful examples nor a missing counterexample in a short list establishes a claim over an entire domain.
Check Your Understanding
Use the conditional truth table and the equivalences proved here to justify your answers.
- State the hypothesis and conclusion in “\(q\) whenever \(p\)” and “\(p\) only if \(q\).” Write each sentence symbolically.
- If \(p\to q\) is true and \(q\) is false, what must the value of \(p\) be? If \(p\to q\) is true and \(p\) is false, what can you conclude about \(q\)?
- Rewrite \((p\lor q)\to r\) without an implication symbol, using De Morgan's Laws where appropriate. Then write an equivalent formula for its negation.
- At a specified real \(x\), negate “If \(x\geq0\), then \(x>2\).” Evaluate both the conditional and its negation at \(x=-1\), \(x=0\), and \(x=2\).
- Disprove the claim “For every real \(x\), if \(x>1\), then \(x>3\).” Explain why \(x=0\) is not a counterexample, and give an input that is.
- Explain why \((x>5)\to(x>1)\) holds for every real \(x\), although \(p\to q\) is not a tautology. Which abstract truth assignment is impossible under this interpretation?