Constructive Dilemma (CD)

The last of the eight rules of inference is constructive dilemma (CD). It reasons by cases: if either of two things happens, and each leads somewhere, then one of the two results follows. It completes the set begun in Modus Ponens and Modus Tollens (MP, MT).

Two lines: a conjunction of two conditionals, and the disjunction of their antecedents.
Rule	From	Infer
CD (constructive dilemma)	$(p \to q) \land (r \to s)$ and $p \lor r$	$q \lor s$

Key idea

CD: from a conjunction of two conditionals and the disjunction of their antecedents, infer the disjunction of their consequents. In this system the two conditionals must be joined in one line first, usually by Conj.

Why it works

If $p \lor r$ is true, at least one of $p$, $r$ is true. If $p$ is, then $p \to q$ gives $q$; if $r$ is, then $r \to s$ gives $s$. Either way, at least one of $q$, $s$ is true, so $q \lor s$ is true. That is two uses of MP, one for each case, packed into one rule.

Conj first, then CD.

Worked examples

Worked example: A dilemma in words

"If it rains, we play cards; if it is sunny, we hike. It will rain or be sunny. So we will play cards or hike." Which rule?

Let $p$: it rains; $q$: we play cards; $r$: it is sunny; $s$: we hike.
Premises: $(p \to q) \land (r \to s)$ and $p \lor r$.

Answer: Constructive dilemma: $q \lor s$.

Worked example: Compound parts

From $(\neg p \to t) \land (q \to (r \land s))$ and $\neg p \lor q$, what does CD give?

The antecedents are $\neg p$ and $q$; their disjunction is line 2.
The consequents are $t$ and $r \land s$.

Answer: $t \lor (r \land s)$

Worked example: When the dilemma collapses

From $(p \to q) \land (r \to q)$ and $p \lor r$, CD gives $q \lor q$. How do you get $q$?

$q \lor q$ says $q$, twice.
The rule of replacement Taut (later) turns $q \lor q$ into $q$.

Answer: CD, then Taut.

Common mistake

Using CD with the two conditionals on separate lines. CD cites exactly two lines: the conjunction of the conditionals and the disjunction. With $p \to q$ on line 1 and $r \to s$ on line 2, first write $(p \to q) \land (r \to s)$ by Conj 1, 2, then apply CD. Also check that the disjunction joins the antecedents, not the consequents.

Why bother with CD when two uses of MP would do? Because MP needs the antecedent itself, and in a dilemma you do not know which antecedent is true, only that one of them is. CD is exactly the rule for that situation: reasoning by cases when the cases come as a disjunction.

Some textbooks also have a destructive dilemma (from the conditionals and $\neg q \lor \neg s$, infer $\neg p \lor \neg r$). Our system leaves it out; you can get the same result with other rules.

CD finishes the list of rules of inference: MP, MT, HS, DS, CD, Simp, Conj and Add. All eight share one restriction, which the next lesson, Matching a Rule to Whole Lines, makes precise: they apply only to whole lines.

Constructive Dilemma (CD)

Worked examples

Practice

Builds on

Leads to

Sources and further reading