Mathematical-Reasoning · Unit 3 · Video 2 · Interactive Practice

Only If Points to the Conclusion: Truth Tables, the Biconditional and Tautologies

IKey Formulas

Formula	Name	Meaning
$P \to Q$	Conditional	If $P$ , then $Q$ ; $P$ only if $Q$ ; $Q$ if $P$ ; $Q$ is necessary for $P$ ; $P$ is sufficient for $Q$
$2^n$ rows	Truth table size	$n$ simple statements; columns chosen by working backward from the last operation
$P \leftrightarrow Q$ means $(Q \to P) \wedge (P \to Q)$	Biconditional	True exactly when $P$ and $Q$ have the same truth value
$P \vee \neg P$ and $P \wedge \neg P$	Tautology and contradiction	True in every row, and false in every row

Key Insight: The word if marks the hypothesis wherever it sits, and only if marks the conclusion: $P$ only if $Q$ rules out exactly the row $P$ true, $Q$ false.

IIWhich Way Does the Arrow Point?

Every English form of a conditional has one hypothesis and one conclusion, even when the conclusion is spoken first.

Sentence

IIIBuilding a Table One Column at a Time

Working backward from the last operation fixes the columns; filling them forward settles every row of $2^n$ .

Statement Step 1

IVSame Truth Table, Same Claim

Two statements share a truth table exactly when the biconditional joining them is a tautology.

A B

💡 Challenge: find a statement built only from $\neg$ , $\wedge$ and $\vee$ whose truth table matches $P \to Q$ .

VQuiz Questions

Problem 1 · Rewrite an Only-If Sentence

Given: "A triangle is equilateral only if it is isosceles." Which if-then sentence makes the same claim?

✅ Correct! "Only if" marks the conclusion, so being isosceles is the conclusion and being equilateral is the hypothesis.

❌ Not quite. That reverses the arrow. "

P

only if

Q

" means

P \to Q

, and here

P

is "equilateral", the phrase spoken first.

❌ Not quite. "Only if" gives one direction. The biconditional also claims every isosceles triangle is equilateral, which the sentence never says (and which is false).

❌ Not quite. "Only if" is not "and". The conjunction claims the triangle is both equilateral and isosceles, so it is false for a triangle that is neither. The original sentence rules out only an equilateral triangle that is not isosceles, so it is true for that triangle.

❌ Not quite. Identify which phrase "only if" attaches to; that phrase is the conclusion.

Show solution

Let $P$ : "the triangle is equilateral" and $Q$ : "the triangle is isosceles". The sentence has the form $P$ only if $Q$ .

$P$ only if $Q$ says $P$ cannot be true unless $Q$ is true. That rules out the single situation $P$ true, $Q$ false, which is exactly the one row where $P \to Q$ is false. So the sentence means $P \to Q$ :

If a triangle is equilateral, then it is isosceles.

Problem 2 · The Odd One Out

Given: the conditional $P \to Q$ . Which sentence does NOT mean $P \to Q$ ?

✅ Correct! "

Q

only if

P

" makes

P

the conclusion, so it means

Q \to P

, the converse.

❌ Not quite. The sufficient condition sits at the tail of the arrow:

P

sufficient for

Q

P \to Q

❌ Not quite. The necessary condition sits at the head of the arrow: you cannot have

P

without

Q

, which is

P \to Q

❌ Not quite. "If" marks the hypothesis even when it comes last, so "

Q

P

" has hypothesis

P

: it is

P \to Q

❌ Not quite. Find the hypothesis in each sentence.

Show solution

Read off the hypothesis of each sentence:

$P$ is sufficient for $Q$ : showing $P$ is enough to get $Q$ , so $P \to Q$ .
$Q$ is necessary for $P$ : $P$ cannot hold without $Q$ , so $P \to Q$ .
$Q$ if $P$ : "if" marks the hypothesis $P$ , so $P \to Q$ .
$Q$ only if $P$ : "only if" marks the conclusion $P$ , so $Q \to P$ .

$Q \to P$ differs from $P \to Q$ in the rows $P = T, Q = F$ and $P = F, Q = T$ , so d is the different statement.

Problem 3 · Count the False Rows

Given: the statement $(P \vee \neg R) \to Q$ . How many of the $8$ rows of its truth table make it false?

✅ Correct! The hypothesis

P \vee \neg R

is true in

3

of the

4

rows where

Q

is false.

❌ Not quite. In the video's

(P \wedge \neg Q) \to R

the hypothesis was a conjunction, true in only two rows. Here the hypothesis is a disjunction, which is false only when both parts are false.

❌ Not quite.

4

counts every row with

Q

false. A conditional with a false conclusion is still true when its hypothesis is false too.

❌ Not quite.

6

counts the rows where the hypothesis

P \vee \neg R

is true. The conditional is false only when, in addition,

Q

is false.

❌ Not quite. The conditional is false only when its hypothesis is true and its conclusion is false.

Show solution

Columns in build order: $P$ , $Q$ , $R$ (step 1); $\neg R$ (step 2); $P \vee \neg R$ (step 3); $(P \vee \neg R) \to Q$ (step 4).

Row	$P$	$Q$	$R$	$\neg R$	$P \vee \neg R$	$(P \vee \neg R) \to Q$
1	T	T	T	F	T	T
2	T	T	F	T	T	T
3	T	F	T	F	T	F
4	T	F	F	T	T	F
5	F	T	T	F	F	T
6	F	T	F	T	T	T
7	F	F	T	F	F	T
8	F	F	F	T	T	F

The conditional is false exactly in rows $3$ , $4$ and $8$ : the rows with hypothesis T and conclusion $Q$ = F. That is 3 rows.

Problem 4 · Tautology, Contradiction or Neither

Classify each statement as a tautology, a contradiction, or neither.

(i) $(P \wedge Q) \to P$

(ii) $P \leftrightarrow \neg P$

(iii) $(P \to Q) \to (Q \to P)$

✅ Correct! (i) is true in every row, (ii) is false in every row, and (iii) is false only when

P = F

and

Q = T

❌ Check (i). The conditional could fail only where

P \wedge Q

is true and

P

is false. Can

P \wedge Q

be true while

P

is false?

❌ Check (ii). A biconditional is true exactly when its two sides have the same truth value. Can

P

and

\neg P

ever match?

❌ Check (iii). Try the row

P = F

Q = T

: what are

P \to Q

and

Q \to P

there? Then try

P = T

Q = T

Show solution

(i) $(P \wedge Q) \to P$ is false only if $P \wedge Q$ is T and $P$ is F. But $P \wedge Q$ is T only when $P$ is T, so no such row exists. The column reads T, T, T, T: a tautology.

(ii) $P$ and $\neg P$ always have opposite truth values, and $P \leftrightarrow \neg P$ is true only when the two sides match. The column reads F, F: a contradiction.

(iii) Rows in the order $(P, Q) = (T,T), (T,F), (F,T), (F,F)$ :

$P$	$Q$	$P \to Q$	$Q \to P$	$(P \to Q) \to (Q \to P)$
T	T	T	T	T
T	F	F	T	T
F	T	T	F	F
F	F	T	T	T

The last column has both T and F, so the statement is neither: a conditional does not imply its converse.

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