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A dilemma is a reasoning either of the form

(1) if A then B (2) if C then D (3) A or C (4) therefore, B or D

or of the form

(1) if A then B (2) if C then D (3) not B or not D (4) therefore not A or not C

Are there situations in mathematics where a conclusion of the form (X or Y) or of the forme ( not X or not Y) would be interesting? Are there situations in mathematics in which the dilemma argument form would be usefull? Are there classical examples of mathematical proofs using dilemma?

  • Every disjunction can be rewritten as an implication: if I prove (say) that either the Goldbach Conjecture holds or the Twin Prime Conjecture holds, then I've proved that the failure of Goldbach implies the success of twin prime. Implications "sound better" in many cases, but there's no real difference between implications and disjunctions (= dilemmas) except phrasing (well, in classical mathematics, at least). – Noah Schweber Apr 12 '19 at 18:36

2 Answers2

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p: There exists two irrational $a,b$ such that $a^b$ is rational.

q: $\sqrt{2}^\sqrt{2}$ is rational

r: $\left (\sqrt{2}^\sqrt{2} \right)^\sqrt{2}$ is rational

q $\Rightarrow$ p

($\lnot$ q) $\Rightarrow$ (r $\Rightarrow$ p)

q or ($\lnot$q) therefore p.

Jonas Gomes
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  • -Is the conclusion a disjunction? –  Apr 12 '19 at 15:05
  • The conclusion is $p$, because $r$ is true. In my example $A=q$, $B=p$, $C = \lnot p$ and $D = (r\Rightarrow p)$ – Jonas Gomes Apr 12 '19 at 15:25
  • +1. To clarify @EleonoreSaintJames: This is an example of a mathematical proof using a dilemma (per your final sentence). Basically, we want to show $\exists x(P(x))$, and we've done this by finding specific $A$ and $B$ such that we can prove the dilemma $P(A)\vee P(B)$, which implies the statement $\exists x(P(x))$ whose proof was our original goal. – Noah Schweber Apr 12 '19 at 18:33
  • @Noam Schweber. Thanks for your explanation, it is clearer now! –  Apr 12 '19 at 20:07
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In tha following proof, a constructive dilemma is used in a subordinate derivation.

The subordinate derivation is :

(1) a is the first element OR b is the first element

(2) if a is the first element, then a < b

(2) if b is the first element then b < a

(3) Therefore : a < b OR b < a

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