Deduction, Daniel Bonevac, Blackwell, 2nd etition, 2003, 8.3, problem 4, p. 238

Just how convincing is the following argument:

There is at most one God. So, there are at most two Gods.

The pattern for sentences with 'at most n things with a certain property' is to begin with n + 1 universal quantifiers, and then qualify our statement by saying that if all those things have the property in question, then two of them must be the same. So, 'there is at most one God' means that if 'x' is a God and 'y' is a God, then 'x' is identical to 'y'. So they collapse down into one. In the same way, three collapse down into two, and so on.

1. (x)(y)[(Gx • Gy) ⊃x = y]
2. ∴(x)(y)(z)[(Gx • Gy • Gz) ⊃(x = y ∨ x = z ∨ y = z)]
3. * ¬ (x)(y)(z)[(Gx • Gy • Gz) ⊃(x = y ∨ x = z ∨ y = z)] ......... AIP
4. * (∃x)(∃y)(∃z)[Gx • Gy • Gz • ¬ (x = y) • ¬ (x = z) • ¬ (y = z)] ......... 3 QC
5. * (∃y)(∃z)[Ga • Gy • Gz • ¬ (a = y) • ¬ (a = z) • ¬ (y = z)] ......... 4 EI x/a
6. * (∃z)[Ga • Gm • Gz • ¬ (a = m) • ¬ (a = z) • ¬ (m = z)] ......... 5 EI y/m
7. * Ga • Gm • Gr • ¬ (a = m) • ¬ (a = r) • ¬ (m = r) ......... 6 EI z/r
8. * (y)[(Ga • Gy) ⊃a = y] ......... 1 UI x/a
9. * (Ga • Gm) ⊃a = m ......... 8 UI y/m
10. * Ga • Gm ......... 7 Simp.
11. * a = m ......... 9,10 MP
12. * ¬ (a = m) ......... 6 Simp.
13. * (a = m) • ¬ (a = m) ......... 11,12 Conj.
14. ¬ ¬ (x)(y)(z)[(Gx • Gy • Gz) ⊃(x = y ∨ x = z ∨ y = z)] ......... 3-13 IP
15. (x)(y)(z)[(Gx • Gy • Gz) ⊃(x = y ∨ x = z ∨ y = z)] ......... 14 DN