The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.4E, 3(d), p. 556

The solution to this problem is most likely on the Students' Solutions CD that comes with the book but the problem has caught my eye in the exercise sets pages and made me attempt my own. The task: prove that the conclusion follows.

1. (∃x)Cx ⊃ Ch
2. ∴(∃x)Cx ≡ Ch
3. * ¬ [(∃x)Cx ≡ Ch] ......... AIP
4. * ¬ {[(∃x)Cx ⊃ Ch] • [Ch ⊃ (∃x)Cx]} ......... 3 BE
5. * ¬ [(∃x)Cx ⊃ Ch] ∨ ¬ [Ch ⊃ (∃x)Cx] ......... 4 DeM
6. * ¬ [Ch ⊃ (∃x)Cx] ......... 1,5 DS
7. * ¬ [¬ Ch ∨ (∃x)Cx] ......... 6 MI
8. * Ch • ¬ (∃x)Cx ......... 7 DeM
9. * Ch ......... 8 Simp.
10. * (∃x) Cx ......... 9 EG
11. * ¬ (∃x)Cx ......... 8 Simp.
12. * (∃x) Cx • ¬ (∃x)Cx ......... 10,11 Conj.
13. ¬ ¬ [(∃x)Cx ≡ Ch] ......... 3-12 IP
14. (∃x)Cx ≡ Ch ......... 13 DN