Constructive dilemma

From Wikipedia, the free encyclopedia
Jump to: navigation, search

Constructive dilemma[1][2][3] is a name of a valid rule of inference of propositional logic. It is the inference that, if P implies Q and R implies S and either P or R is true, then Q or S has to be true. In sum, if two conditionals are true and at least one of their antecedents is, then at least one of their consequents must be too. Constructive dilemma is the disjunctive version of modus ponens, whereas, destructive dilemma is the disjunctive version of modus tollens. The rule can be stated:

\frac{P \to Q, R \to S, P \or R}{\therefore Q \or S}

where the rule is that whenever instances of "P \to Q", "R \to S", and "P \or R" appear on lines of a proof, "Q \or S" can be placed on a subsequent line.

Formal notation[edit]

The constructive dilemma rule may be written in sequent notation:

(P \to Q), (R \to S), (P \or R) \vdash (Q \or S)

where \vdash is a metalogical symbol meaning that Q \or S is a syntactic consequence of P \to Q, R \to S, and Q \or S in some logical system;

and expressed as a truth-functional tautology or theorem of propositional logic:

(((P \to Q) \and (R \to S)) \and (P \or R)) \to (Q \or S)

where P, Q, R and S are propositions expressed in some formal system.

Variable English[edit]

If P then Q. If R then S. P or R. Therefore, Q or S.

Natural language example[edit]

If I win a million dollars, I will donate it to an orphanage.
If my friend wins a million dollars, he will donate it to a wildlife fund.
I win a million dollars or my friend wins a million dollars.
Therefore, either an orphanage will get a million dollars, or a wildlife fund will get a million dollars.

The dilemma derives its name because of the transfer of disjunctive operator.

References[edit]

  1. ^ Hurley, Patrick. A Concise Introduction to Logic With Ilrn Printed Access Card. Wadsworth Pub Co, 2008. Page 361
  2. ^ Moore and Parker
  3. ^ Copi and Cohen