// laws and rules · Rule of inference
Constructive dilemma_
The constructive dilemma is a four-variable rule of inference: from p ⇒ q, r ⇒ s and p ∨ r we conclude q ∨ s. It is a modus ponens “in parallel”: we know p or r happens, and each has its consequence, so one of the consequences must happen. The full formula is a 16-row tautology.
Example
(((p ⇒ q) ∧ (r ⇒ s)) ∧ (p ∨ r)) ⇒ (q ∨ s)
What the variables mean
- ▸ p: “I take the train”
- ▸ q: “I arrive at 8”
- ▸ r: “I take the bus”
- ▸ s: “I arrive at 9”
In plain words
If I take the train I arrive at 8; if I take the bus I arrive at 9. I will take the train or the bus. Therefore, I will arrive at 8 or at 9.
Truth table
| p | q | r | s | p ⇒ q | r ⇒ s | (p ⇒ q) ∧ (r ⇒ s) | p ∨ r | ((p ⇒ q) ∧ (r ⇒ s)) ∧ (p ∨ r) | q ∨ s | (((p ⇒ q) ∧ (r ⇒ s)) ∧ (p ∨ r)) ⇒ (q ∨ s)★ |
|---|---|---|---|---|---|---|---|---|---|---|
| T | T | T | T | T | T | T | T | T | T | T |
| T | T | T | F | T | F | F | T | F | T | T |
| T | T | F | T | T | T | T | T | T | T | T |
| T | T | F | F | T | T | T | T | T | T | T |
| T | F | T | T | F | T | F | T | F | T | T |
| T | F | T | F | F | F | F | T | F | F | T |
| T | F | F | T | F | T | F | T | F | T | T |
| T | F | F | F | F | T | F | T | F | F | T |
| F | T | T | T | T | T | T | T | T | T | T |
| F | T | T | F | T | F | F | T | F | T | T |
| F | T | F | T | T | T | T | F | F | T | T |
| F | T | F | F | T | T | T | F | F | T | T |
| F | F | T | T | T | T | T | T | T | T | T |
| F | F | T | F | T | F | F | T | F | F | T |
| F | F | F | T | T | T | T | F | F | T | T |
| F | F | F | F | T | T | T | F | F | F | T |
Classification: Tautology · 16 rows
Statement of the rule
Premises: p ⇒ q, r ⇒ s and p ∨ r. Conclusion: q ∨ s. With four variables the table has 2⁴ = 16 rows.
It is called a “dilemma” because the disjunction p ∨ r presents two paths, and “constructive” because it builds an affirmative conclusion from them (unlike the destructive dilemma, which concludes ¬p ∨ ¬r from ¬q ∨ ¬s).
Why it is valid: reading the table
The conclusion q ∨ s is F only when q = F and s = F. In those four rows, for p ∨ r to be T we need p = T or r = T. If p = T then p ⇒ q is F (since q = F); if r = T then r ⇒ s is F (since s = F). Either way some premise fails, the conjunction is F and the final implication is T.
In the remaining twelve rows the conclusion q ∨ s is already T, so the implication is automatically T. No counterexample exists.
How it is used in proofs
It is the formal version of proof by cases: if we know A or B holds and each case leads to a conclusion, we may assert the disjunction of the conclusions without knowing which case actually occurs.
It is derived from two applications of modus ponens inside a disjunction elimination.
Examples
Everyday: “If I study medicine I will be a doctor; if I study law I will be a lawyer. I will study medicine or law. Therefore I will be a doctor or a lawyer.”
Programming: “If the cache hits, the response takes 5 ms; if it misses, it takes 50 ms. Every request either hits or misses. So every request takes 5 ms or 50 ms.” This is the usual reasoning when bounding running time by cases.
Relation to other laws
When p and r are the same proposition and so are q and s, the constructive dilemma collapses into modus ponens. When q and s coincide we get the “proof by cases” rule: from p ⇒ q, r ⇒ q and p ∨ r, conclude q.
The destructive dilemma is its contrapositive: from p ⇒ q, r ⇒ s and ¬q ∨ ¬s conclude ¬p ∨ ¬r, justified in the same way via modus tollens.
Try it yourself
Edit the expression in the calculator and watch how every step of the table changes.
Open in the calculator →Related operators
Frequently asked questions
Why does it have 16 rows? ▼
Because four variables (p, q, r, s) are involved and 2⁴ = 16.
What is the destructive dilemma? ▼
The mirror rule: from p ⇒ q, r ⇒ s and ¬q ∨ ¬s conclude ¬p ∨ ¬r. It uses modus tollens instead of modus ponens.
Can I conclude q ∧ s instead of q ∨ s? ▼
No. We only know that p or r happens, not both, so we can only guarantee one of the consequences.