Constructive dilemma

Constructive dilemma

In logic, a constructive dilemma is a formal logical argument that takes the form:

1a) P → Q.
b) R → S.
2) Either P or R is true.

Therefore, either Q or S is true.

In logical operator notation with three premises

 P \rightarrow Q
 R \rightarrow S
 P \lor R
 \therefore Q \lor S .

In logical operator notation with two premises[1]

 ( P \rightarrow Q ) \land ( R \rightarrow S )
 P \lor R
 \therefore Q \lor S .

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.

An example:

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.
Either 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 operants.

Proof

1.  (P \rightarrow Q) \and (R \rightarrow S) \and (P \vee R))
2.  \Leftrightarrow (\neg P \vee Q) \and (\neg R \vee S) \and (P \vee R)
3.  \Rightarrow (\neg P \vee (Q \vee S)) \and (\neg R \vee (Q \vee S)) \and (P \vee R) (addition)
4.  \Rightarrow (\neg P \vee (Q \vee S)) \and (\neg R \vee (Q \vee S)) (simplification)
5.  \Leftrightarrow (Q \vee S) \vee (\neg P \and \neg R) (distribution)
6.  \Leftrightarrow (Q \vee S) \vee \neg(P \vee R) (DeMorgan's Law)
7.  \Leftrightarrow (Q \vee S) \vee False (from assumption)
7.  \Leftrightarrow Q \vee S

References

  1. ^ Hurley, Patrick. A Concise Introduction to Logic With Ilrn Printed Access Card. Wadsworth Pub Co, 2008. Page 361