Editing Constructive dilemma (section)

{{Short description|Rule of inference of propositional logic}}
{{Infobox mathematical statement
| name = Constructive dilemma
| type = [[Rule of inference]]
| field = [[Propositional calculus]]
| statement = If <math>P</math> implies <math>Q</math> and <math>R</math> implies <math>S</math>, and either <math>P</math> or <math>R</math> is true, then either <math>Q</math> or <math>S</math> has to be true.
| symbolic statement = <math>\frac{(P \to Q), (R \to S), P \lor R}{\therefore Q \lor S}</math>
}}
{{Transformation rules}}

'''Constructive dilemma'''<ref>Hurley, Patrick. A Concise Introduction to Logic With Ilrn Printed Access Card. Wadsworth Pub Co, 2008. Page 361</ref><ref>Moore and Parker</ref><ref>Copi and Cohen</ref> is a [[Validity (logic)|valid]] [[rule of inference]] of [[propositional calculus|propositional logic]]. It is the [[inference]] that, if ''P'' implies ''Q'' and ''R'' implies ''S'' and either ''P'' or ''R'' is true, then either ''Q or S'' has to be true. In sum, if two [[material conditional|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 [[Logical disjunction|disjunctive]] version of [[modus ponens]], whereas [[destructive dilemma]] is the disjunctive version of ''[[modus tollens]]''. The constructive dilemma rule can be stated:

:<math>\frac{(P \to Q), (R \to S), P \lor R}{\therefore Q \lor S}</math>

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