I think I solved it. Since (by axiom group 2 of Enderton's book and modus ponens) and by assumption, which means , i.e., inconsistent.

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- Nov 10th 2011, 06:55 AM #1

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## Fol 6

Enderton problem 4 in Section 2.5

Let . Is consistent? Is satisfiable?

The completeness theorem says that any consistent set of formulas is satisfiable. Therefore, we only need to show that is consistent.

Suppose to the contrary, towards a contradiction, . Then, we have both and for any wff . I am looking for a counterexample , which is either or , not both.

Any help will be appreciated.

- Nov 10th 2011, 07:18 AM #2

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- Nov 10th 2011, 08:10 AM #3

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