The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, Ex. 10.5E, 4(f), p. 563

Show that the following sentences are equivalent. Solution: if they are equivalent, then we can derive one from the other.

(x)[Ax • (∃y) ¬ Bxy] is equivalent to ¬ (∃x)[¬ Ax ∨(y)(Bxy • Bxy)]

1. (x)[Ax • (∃y) ¬ Bxy]
2. * ¬ ¬ (∃x)[¬ Ax ∨(y)(Bxy • Bxy)] / AIP
3. * (∃x)[¬ Ax ∨(y)(Bxy • Bxy)] / 2DN
4. * ¬ Aa ∨(y)(Bay • Bay) / 3EI x/a
5. * Aa • (∃y) ¬ Bay / 1UI x/a
6. * Aa / 5Simp.
7. * (y)(Bay • Bay) / 4,6DS
8. * (∃y) ¬ Bay / 5Simp.
9. * ¬ Bam / 8EI y/m
10. * Bam • Bam / 7UI y/m
11. * Bam / 10Taut.
12. * ¬ Bam • Bam / 9,11Conj.
13. * ¬ ¬ ¬ (∃x)[¬ Ax ∨(y)(Bxy • Bxy)] / 2-12IP
14. * ¬ (∃x)[¬ Ax ∨(y)(Bxy • Bxy)] / 13DN

We have shown that the second sentence is derivable from the first by indirect proof.

1. ¬ (∃x)[¬ Ax ∨(y)(Bxy • Bxy)]
2. (x)[Ax • ¬ (y)(Bxy • Bxy)] / 1QC
3. Ax • ¬ (y)(Bxy • Bxy) / 2UI x/x
4. Ax / 3Simp.
5. ¬ (y)(Bxy • Bxy) / 3Simp.
6. (∃y)(¬ Bxy ∨¬ Bxy) / 5DeM
7. ¬ Bxm ∨¬ Bxm / 6EI y/m
8. ¬ Bxm / 7Taut.
9. (∃y)¬ Bxy / 8EG
10. Ax • (∃y)¬ Bxy / 4,9Conj.
11. (x)[Ax • (∃y)¬ Bxy ] / 10UG

The first sentence is also derivable from the second. So, the two sentences are equivalent.