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What is wrong with this proof?
Theorem. 3 is less than 1.
Proof. Every number is either less than 1 or greater than 1 or equals 1. Let $c$ be an arbitrary number. Therefore, it is less than 1 or greater than 1 or equals 1. Suppose it is less than 1. By the rule of universal generalization, if an arbitrary number is less than 1, every number is less than 1. Therefore, 3 is less than 1.

Gerry Myerson
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user182280
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    The rule of generalization applies when you make no assumptions whatsoever over the variable, when it's a free variable, wich is not the case as you said "suppose it is less than 1" – Jonas Gomes Oct 10 '14 at 03:24
  • The rule of generalization, namely, if $\Gamma \vdash \phi(x)$, then $\Gamma \vdash \forall x \phi(x)$ actually requires that $x$ is a free variable. – Jonas Gomes Oct 10 '14 at 03:26
  • Why would you use something as powerful as the rule of universal generalization here? That applies to strictly formal arguements where variables have no assumed values and clearly that's not the case here! I get you want to practice your logic skills,but you have to be very careful with actual mathematical arguements since they have many more assumptions then strictly logical ones. – Mathemagician1234 Oct 10 '14 at 03:33
  • ...what do you think is wrong with the proof? – DanielV Oct 10 '14 at 07:48

2 Answers2

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You are fine up to "Suppose it is less than 1" You need to review your rule of universal generalization (which you did not quote), which will likely fail here.

Ross Millikan
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Once you've decided $c$ is less than $1$, it's not arbitrary anymore.

Neal
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