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The problem is stated correctly. You are reading it incorrectly, but I am not sure how. If c=0, then it says that f(x)=f(y) if and only if (<=>) x=y. In othe
by greeneggs 6y ago
The problem is stated correctly. You are reading it incorrectly, but I am not sure how.
If c=0, then it says that f(x)=f(y) if and only if (<=>) x=y. In other words, for all x≠y, f(x)≠f(y). Thus the function is 1-to-1. This is a nontrivial property that is difficult to check.
- codebje 6y agoThe incorrect reading is that x and y are free. They're universally quantified, as you've said in your comment.
- Gehinnn 6y agoThe first step you do when proving something all-quantified is making the quantified variables free.
- codebje 6y agoCould you explain that a little for me? Do you mean substituting a term to show a counter-example?
- Gehinnn 6y agoIf you want to prove "\forall x: A(x)", you start with "let x be a free variable. It remains to show A(x)".
- codebje 6y agoWouldn't that only show \exists x: A(x)? Or perhaps by free variable you mean one whose value isn't decidable - but that contradicts having chosen values in the origin comment.