Saying a conditional without if-then
You have rules for “and”, “or” and “not”. Is “if ... then ...” really a fourth kind of statement, or can it be built out of the other three?
Where you are 3 in Logical equivalence
Why this exists
“And”, “or” and “not” come with an algebra: De Morgan’s laws push a negation through them, and the disjunctive syllogism lets you eliminate part of an “or”. A conditional, “if P then Q”, looks like it stands outside that algebra entirely, a fourth and stranger kind of statement with its own private rules.
It is not. “If P then Q” says exactly what “P is false, or Q is true” says, no more and no less, so every tool already built for “or” and “not” works on a conditional as soon as it is written that way. The single biggest payoff is negation. Negating an “if ... then ...” head-on is confusing, since it is not obvious what the opposite of a promise even means. Negating an “or” is routine: push the negation through with De Morgan’s laws. Rewrite the conditional first, and its negation stops being a special case.
Definition
Let P and Q be statements. The conditional P \to Q (“if P then Q”) is logically equivalent to the disjunction \neg P \lor Q:
Two statements are logically equivalent, the subject of logical equivalence, when they have the same truth value for every assignment of true and false to P and Q. \neg P \lor Q reads two ways that both match “if P then Q” in plain English:
- “P is false, or Q is true.”
- “Either not P, or Q.”
Both readings use only ”or” and negation, the two connectives this rewriting needs. The conditional itself and its truth table are covered in the truth table for the conditional, and the words converse, inverse and contrapositive that go with it in contrapositive equivalence; this lesson does not repeat them, only puts the conditional in a new set of clothes.
Intuition
A conditional P \to Q breaks in exactly one way: P true and Q false. Every other combination leaves it standing, including the case where P never holds at all.
Look at \neg P \lor Q with that in mind. It fails only when both of its parts fail: \neg P false, meaning P is true, and Q false. That is the same single case. Read it as a promise with two ways to keep it: either the condition P never came up (\neg P is true), so there was nothing to deliver on, or the condition did come up and Q came through anyway. The only way to break the promise is for P to come up and Q to fail, and that is exactly what the disjunction checks.
This is also why a conditional with a false hypothesis is automatically true, a fact usually called vacuous truth. If P is false, then \neg P is already true, and a disjunction with one true part is true no matter what its other part says. “If P then Q” is not claiming Q; it is only claiming that P never happens without Q, and a P that never happens keeps that claim for free.
Results
Implication as disjunction. For any statements P and Q,
Negation of a conditional. Negating both sides, then pushing the negation through the “or” on the right with De Morgan’s law, gives
The negation of a conditional is a conjunction, never another conditional. To show “if P then Q” is false, you must produce a single case where P is true and Q is false; nothing short of that will do.
Why it is true
A truth table settles it directly. Fill in P \to Q from its own definition (false only when P is true and Q is false) and \neg P \lor Q from the definition of “or”, with T for true and F for false:
| Row | P | Q | \neg P | P \to Q | \neg P \lor Q |
|---|---|---|---|---|---|
| 1 | T | T | F | T | T |
| 2 | T | F | F | F | F |
| 3 | F | T | T | T | T |
| 4 | F | F | T | T | T |
The last two columns match in every row, so the two statements are logically equivalent.
The negation follows by pushing a negation through the equivalent form:
The middle step is De Morgan’s law for “or”: negating a disjunction negates each part and swaps “or” for “and”. The last step cancels a double negation: “it is not the case that P is false” says exactly P.
Worked examples
Rewriting a conditional as an or-statement
Rewrite “If a number is a multiple of 8, then it is a multiple of 4” without using the word “if”.
SolutionHere P is “a number is a multiple of 8” and Q is “a number is a multiple of 4”. By the equivalence, the statement is the same as \neg P \lor Q:
“A number is not a multiple of 8, or it is a multiple of 4.”
Every multiple of 8 is 8k for some integer k, and 8k = 4(2k) is a multiple of 4, so the statement is true for every number, and so is its or-form.
Reading a conditional true because the hypothesis fails
Let P be “12 is a multiple of 4” and Q be “12 is even”. Let R be “7 is a multiple of 4” and S be “7 is even”. Find the truth value of P \to Q and \neg P \lor Q, then of R \to S and \neg R \lor S.
SolutionP is true and Q is true, so P \to Q is true, and \neg P \lor Q = \text{F} \lor \text{T} is also true.
R is false (7 is not a multiple of 4) and S is false (7 is odd). Since the hypothesis R is false, R \to S is vacuously true, and indeed \neg R \lor S = \text{T} \lor \text{F} is true as well: \neg R alone already carries it, and S is not even needed.
Negating a conditional
Negate “If the alarm sounds, the doors lock” without writing another conditional.
SolutionLet P be “the alarm sounds” and Q be “the doors lock”. First rewrite as an or-statement: P \to Q is \neg P \lor Q, “the alarm does not sound, or the doors lock”. Its negation, by De Morgan’s law for “or”, is the conjunction of the negated parts:
The alarm sounds, and the doors do not lock.
A tempting guess is “if the alarm sounds, the doors do not lock”, but that is still a conditional, P \to \neg Q, and it cannot be the negation of P \to Q. Suppose the alarm never sounds at all, so P is false: then P \to Q is vacuously true, and P \to \neg Q is vacuously true as well, since a false hypothesis carries any conditional. Two statements that are both true in the same case cannot be negations of each other, since a statement and its negation always have opposite truth values.
Disproving a conditional
Show that the claim “for every real number x, if x > 0, then x^2 > 1” is false.
SolutionBy the negation result, the conditional fails at a value of x exactly when x > 0 holds and x^2 > 1 fails there, that is, when x > 0 and x^2 \le 1. A claim about every x is broken by one such value. Take \displaystyle x = \frac{1}{2}:
Both halves of the negation hold at \displaystyle x = \frac{1}{2}, so the original claim is false. Naming the exact shape the negation takes, a hypothesis that holds together with a conclusion that fails, is what turns “find a counterexample” from a hunch into a search with a target.
Proving a conditional by casing on its or-form
Prove that for every integer n, if n is not a multiple of 3, then n^2 - 1 is a multiple of 3.
SolutionLet P be “n is not a multiple of 3” and Q be “n^2 - 1 is a multiple of 3”. The goal P \to Q is the same statement as \neg P \lor Q, “n is a multiple of 3, or n^2 - 1 is a multiple of 3”. Every integer leaves remainder 0, 1 or 2 on division by 3, so it is enough to show one part of the “or” holds in each of those three cases.
If n = 3k for an integer k, the first part is already true, and there is nothing to show.
If n = 3k + 1, then n^2 - 1 = 9k^2 + 6k = 3(3k^2 + 2k), a multiple of 3.
If n = 3k + 2, then n^2 - 1 = 9k^2 + 12k + 3 = 3(3k^2 + 4k + 1), a multiple of 3.
One part of \neg P \lor Q holds in every case, so the conditional is true for every integer. Check. n = 5 gives 24 = 3 \cdot 8, and n = 7 gives 48 = 3 \cdot 16. This is a proof by cases with the goal read as an “or”: the case where \neg P holds costs nothing, and all the work goes into the cases where P holds.
Boundaries
The negation of a conditional is a conjunction, not another conditional. “If P then not Q” is a different statement from the negation of “if P then Q”, as Example 3 shows: both can be vacuously true at once, which a statement and its true negation can never be.
A claim about every x needs its quantifier handled too. “If x > 2, then x^2 > 4” has no truth value until x is given a value or is quantified. The equivalence holds value by value: for each x, the conditional and “x \le 2 or x^2 > 4” agree. But the negation of “for every x, if P then Q” is not “P and not Q” for every x; it is “for some x, P and not Q”, one counterexample, as in Example 4. Turning “every” into “some” is the separate step of negating a quantified statement.
Order is not interchangeable. \neg P \lor Q matches P \to Q. Swapping which part is negated, \neg Q \lor P, matches Q \to P instead, the converse, and a conditional need not agree with its converse. Only swapping and negating both parts together preserves the meaning.
Rewriting is not proving. Turning P \to Q into \neg P \lor Q changes the words, not the truth. A false conditional becomes a false disjunction; the rewriting tells you what a counterexample must look like, but you still have to find or rule one out, as in Examples 4 and 5.
Where this goes next
The negation form is the engine behind proof by contradiction applied to a conditional: to assume the conditional is false is to assume P and \neg Q together, and from there you argue toward something impossible. It is also the reading behind modus tollens, where denying Q in \neg P \lor Q forces \neg P, and behind the contrapositive, a second way of writing the same statement that keeps its meaning exactly.
Read the other way, as “\neg P or Q”, a conditional turns into ordinary “or” reasoning: the disjunctive syllogism applied to \neg P \lor Q and P is exactly what makes modus ponens valid.
Exercises
Rewriting without if-then
- “If a number is a multiple of 12, then it is a multiple of 4.” Rewrite it without “if”.
- “If x > 3, then x^2 > 9.” Rewrite it without “if”, and say whether it holds for every real number x.
- “If today is Tuesday, then tomorrow is Wednesday.” Rewrite it without “if”.
- “If n is a multiple of 9, then n is a multiple of 3.” Rewrite it without “if”, and say whether the whole statement is true for every integer n.
Statements and truth values
Let P be “18 is a multiple of 6” and Q be “18 is odd”. Let R be “18 > 100”.
- Write P \to Q as a disjunction, and find its truth value.
- Write R \to Q as a disjunction, and find its truth value.
- Find the truth value of P \to R.
- Find the truth value of \neg Q \to P.
- Find the truth value of \neg P \to R.
Negating a conditional
- Example 3 negates “If the alarm sounds, the doors lock.” Negate “If the sprinklers turn on, the lawn gets wet” the same way, without writing another conditional.
- Negate “If n is prime, then n is odd” without writing another conditional. Then test the negation at n = 2, and say what this shows about the original statement.
- Negate “If x^2 = 16, then x = 4” without writing another conditional. Then name a value of x that makes the negation true.
- Negate P \to (Q \lor R) without writing another conditional.
Disproving with the negation
- Disprove “If n is a multiple of 4, then n is a multiple of 8” by naming a value of n that makes the hypothesis true and the conclusion false.
- Disprove “If x < 5, then x > 0” by naming a value of x that makes the hypothesis true and the conclusion false.
Reasoning
- Explain why P \to Q and \neg P \lor Q agree in every row where P is false, without checking what Q is.
- A classmate negates “If it snows, school is canceled” as “If it snows, school is not canceled.” Explain why this cannot be the negation, without discussing the weather at all.
- Without a truth table, prove that P \to (Q \to R) and Q \to (P \to R) are logically equivalent, so the order in which two hypotheses are listed never matters.
Answers
- “A number is not a multiple of 12, or it is a multiple of 4.” True for every number, since 12k = 4(3k).
- “x \le 3, or x^2 > 9.” The negation of x > 3 is x \le 3, not x < 3. It holds for every real x: if x > 3, then x^2 - 9 = (x - 3)(x + 3) is a product of two positive numbers, so x^2 > 9.
- “Today is not Tuesday, or tomorrow is Wednesday.”
- “n is not a multiple of 9, or n is a multiple of 3.” True for every integer, since 9k = 3(3k).
- \neg P \lor Q: “18 is not a multiple of 6, or 18 is odd.” P is true and Q is false, so P \to Q is false, and \neg P \lor Q = \text{F} \lor \text{F} is false too.
- \neg R \lor Q: “18 is not more than 100, or 18 is odd.” R is false, so R \to Q is vacuously true, matching \neg R \lor Q = \text{T} \lor \text{F} = \text{true}.
- P is true and R is false, so P \to R is false.
- \neg Q is true (18 is not odd) and P is true, so \neg Q \to P is true.
- \neg P is false (18 is a multiple of 6), so \neg P \to R is vacuously true.
- “The sprinklers turn on, and the lawn does not get wet.”
- “n is prime and n is not odd.” At n = 2: 2 is prime, and 2 is not odd, so the negation is true at n = 2. That means the original statement, “if n is prime, then n is odd”, is false, with n = 2 as the counterexample.
- “x^2 = 16 and x \ne 4.” At x = -4: (-4)^2 = 16 and -4 \ne 4, so the negation is true there.
- P \land \neg Q \land \neg R. De Morgan’s law gives \neg(Q \lor R) \equiv \neg Q \land \neg R, so the whole negation is P \land (\neg Q \land \neg R).
- n = 4. 4 is a multiple of 4, and 4 is not a multiple of 8.
- x = -1. -1 < 5 is true, and -1 > 0 is false.
- When P is false, \neg P is true, and a disjunction with one true part is true regardless of its other part, so \neg P \lor Q is true. When P is false, P \to Q is vacuously true by definition. Both are true in that row no matter what Q is, so they agree without Q ever being checked.
- The original is P \to Q and the guess is P \to \neg Q. In any row where P is false, both are vacuously true at once. A statement and its negation always have opposite truth values, so the guess cannot be the negation. The real negation is the conjunction P \land \neg Q: “it snows, and school is not canceled”.
- Rewrite each arrow as an “or”, inner one first. Q \to R is \neg Q \lor R, so P \to (Q \to R) is \neg P \lor (\neg Q \lor R). In the same way Q \to (P \to R) is \neg Q \lor (\neg P \lor R). An “or” of several parts is true exactly when at least one part is true, whatever the grouping and order, so both are “\neg P or \neg Q or R” and have the same truth value in every row. Each rewriting step was an equivalence, so the two conditionals are logically equivalent.