Symbolic LogicConditional and Indirect Proof

Indirect Proof (IP)

To prove p, assume ¬p, derive a contradiction q ∧ ¬q, then close the subproof and write p.

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Indirect proof (IP), also called proof by contradiction, is the second new method. To show a statement is true, you assume it is false and show that leads to a contradiction. The layout is the same subproof you used in Conditional Proof (CP).

Key idea

To prove $A$ by IP: assume $\neg A$ (justified AIP) in a subproof. Derive a contradiction, a line of the form $q \land \neg q$. Close the subproof and write $A$, justified "IP i–j". (If you assumed $B$ instead, IP gives $\neg B$.)

Why it works

If the premises together with $\neg A$ lead to $q \land \neg q$, which is false in every row, then the premises and $\neg A$ cannot all be true. So whenever the premises are true, $\neg A$ is false, which means $A$ is true. IP is the formal version of "that can't be right, so the opposite must be".
Indirect proof: assume, reach q ∧ ¬q, conclude the negation of the assumption.

Worked examples

Worked example: Prove a letter

Premises $p \lor q$ and $\neg q$. Prove $p$ by IP.
  1. Line 3: $\neg p$ (AIP).
  2. Line 4: $q$ by DS 1, 3.
  3. Line 5: $q \land \neg q$ by Conj 4, 2.
  4. Line 6: $p$ by IP 3–5.

Answer: IP gives the opposite of the assumption $\neg p$, which is $p$.

Worked example: What to assume

You want $\neg (r \land s)$ by IP. What do you assume?
  1. Assume the negation of the goal: $\neg \neg (r \land s)$.
  2. IP also lets you assume $r \land s$ directly, since its negation is the goal.

Answer: $r \land s$ (or $\neg \neg (r \land s)$).

Worked example: Reaching the contradiction

Inside an IP subproof you have $t$ on line 5 and $\neg t$ on line 7. What next?
  1. Join them by Conj.

Answer: $t \land \neg t$ by Conj 5, 7; then close with IP.

Common mistake

Stopping when you reach some line that looks false, such as $\neg p$ when you "know" $p$. The subproof must end in an explicit contradiction of the form $q \land \neg q$, written on one line. If you have $q$ and $\neg q$ on separate lines, use Conj to put them together before you close.
Any contradiction will do; it need not involve the letters of the conclusion.

Notice how IP differs from CP. CP proves a conditional and gives you something to aim for: the consequent. IP can prove anything, but the target inside the subproof is vaguer: any contradiction at all. A good strategy is to look for a formula and its negation that are both within reach, then aim for each of them.

IP is often the easiest way to prove a negation, or a simple letter that appears only deep inside the premises. When nothing else suggests itself, assume the opposite of the conclusion and look for two lines that clash.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • Patrick J. Hurley and Lori Watson, A Concise Introduction to Logic, 13th ed., section 7.6, Indirect Proof
  • forall x: Calgary, an open introduction to formal logic (P. D. Magnus, Tim Button and others; Open Logic Project), chapter on negation and reductio
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