Worksheet

Worksheet: equivalent formulas, negating statements

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Summary

A tautology is a formula that is true on every row of its truth table, and two formulas are equivalent when their columns agree. A compound statement is negated with De Morgan's laws and the negation of an implication. The worksheet trains truth tables, negating statements and the converse, inverse and contrapositive.

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Remember

$A \equiv B$ means that the formulas $A$ and $B$ have the same truth values on every row of the table. A formula that is true on all rows is a tautology.

  1. Complete the table to show that the formula $(p \land (p \rightarrow q)) \rightarrow q$ is a tautology.

    $p$ $q$ $p \rightarrow q$ $p \land (p \rightarrow q)$ $(p \land (p \rightarrow q)) \rightarrow q$
    $1$ $1$
    $1$ $0$
    $0$ $1$
    $0$ $0$

  2. Complete the table to show that $p \land (q \vee r) \equiv (p \land q) \vee (p \land r)$.

    $p$ $q$ $r$ $q \vee r$ $p \land (q \vee r)$ $p \land q$ $p \land r$ $(p \land q) \vee (p \land r)$
    $1$ $1$ $1$
    $1$ $1$ $0$
    $1$ $0$ $1$
    $1$ $0$ $0$
    $0$ $1$ $1$
    $0$ $1$ $0$
    $0$ $0$ $1$
    $0$ $0$ $0$

  3. Write the negation of each statement, using De Morgan's laws or the negation of an implication. State the truth value of the statement and of its negation.

    Statement Negation Values
    a) “$6$ is divisible by $2$ and by $4$.”
    b) “$\pi > 3$ or $\pi < 2$.”
    c) “If $3 > 2$, then $3^{2} > 2^{2}$.”
    d) “$- 1 \in {\mathbb{N}}$ or $( - 1)^{2} \in {\mathbb{N}}$.”

  4. Consider the implication: “If a natural number is divisible by $6$, then it is divisible by $3$.” Write its converse, its inverse and its contrapositive, and state which of the four statements are true.

    Converse: …

    Inverse: …

    Contrapositive: …

    Truth values of the four statements: …

  5. Given that the proposition $(p \land \overline{q}) \rightarrow r$ is false, find $v(p)$, $v(q)$, $v(r)$, then the truth value of the propositions $(p \rightarrow r) \vee q$ and $\overline{q} \leftrightarrow p$.

    a) $v(p)$ = … ; $v(q)$ = … ; $v(r)$ = …

    b) $v((p \rightarrow r) \vee q)$ = …

    c) $v(\overline{q} \leftrightarrow p)$ = …

  6. Show that the formula $(p \rightarrow q) \vee (q \rightarrow p)$ is a tautology. What does this mean for any two propositions?

Frequently asked questions

What is a tautology?

A formula that takes the value $1$ on every row of its truth table, whatever the values of the variables. Simple examples are $p \vee \overline{p}$ and $p \rightarrow p$. To show that a formula is not a tautology, it is enough to find one row where it takes the value $0$.

How do I negate a statement that contains “and” or “or”?

You negate each simple proposition and switch the operator: “and” turns into “or”, and “or” turns into “and” (De Morgan's laws). For example, the negation of “$x > 0$ and $y > 0$” is “$x \leq 0$ or $y \leq 0$”. Relations are negated too: $<$ becomes $\geq$, and $\in$ becomes $\notin$.

Is the inverse of an implication the same as its negation?

No. The inverse of $p \rightarrow q$ is still an implication, $\overline{p} \rightarrow \overline{q}$, while the negation is $p \land \overline{q}$. The inverse always has the same truth value as the converse, not as the original implication.