An extension of section 16.4.3: Exactly n
Recall that "exactly one" can be broken down into "at least one" and "at most one". Thus (1) should be symbolized as (2):
(1) Exactly one student left.
(2) (∃x) (S1x & L1x & (∀y) ((S1y & L1y) ⊃ x = y))
The same principle applies to all sentences containing "exactlyn", no matter how big n is. To symbolize a sentence like (3), we decompose into an "at least" claim, and an "at most" claim.
(3) Exactly two people are tall.
(3) asserts that at least two people are tall, so its symbolization should include "(∃x) (P1x & T1x & (∃y) (P1y & T1y & x ≠y))". (Remember that an "at least n" sentence requires n existentials, and a claim that none of the existentially quantified variables are identical.) (3) also asserts that at most two people are tall, so we need to add that anyone who is tall must be one of these two people. Thus (3) is correctly symbolized as (4):
(4) (∃x) (P1x & T1x & (∃y) (P1y & T1y & x ≠y & (∀z) ((P1z & T1z) ⊃ (z = x v z = y))))
In general, a statement of the form "there are exactly n ⍺" should be symbolized as follows:
Exactly n Procedure:
(∃x1) (⍺1 & … & (∃xn) (⍺ xn & x1≠x2 & … & x1≠xn & x2≠x3 & …. & xn-1≠xn & (∀xn+1) (⍺xn+1 ⊃ (xn+1 = x1 v … v xn+1 = xn)))…)
- There are exactly two frogs. (F1: is frog)
- Exactly two frogs hopped. (H1: hopped)
- Exactly three students passed. (S1: is a student; P1: passed)
- There are exactly three students and they passed.
- Exactly two boys have turtles. (B1: is a boy; T1: is a turtle: H2: has)
- Every boy has exactly two turtles.
- (∃x) (F1x & (∃y) (F1y & x ≠y & (∀z) (F1z ⊃ (z = x v z = y)))
- (∃x) (F1x & H1x & (∃y) (F1y & H1y & x≠y & (∀z) ((F1z & H1z) ⊃ (z = x v z = y))))
- (∃x) (S1x & P1x & (∃y) (S1y & P1y & (∃z) (S1z & P1z & x≠y & y≠z & x≠z & (∀x1) ((S1x1 & P1x1) ⊃ (x1=x v x1=y v x1=z))))
- (∃x) (S1x & (∃y) (S1y & (∃z) (S1z & (∀x1) (S1x1 ⊃ (x1=x v x1=y v x1=z) & P1x & P1y & P1z)))
- (∃x) (B1x & (∃x1) (T1x1 & H2xx1) & (∃y) (B1y & (∃y1) (T1y1 & H2yy1 & x≠y & (∀z) ((B1z & (∃z1) (T1z1 & H2zz1)) ⊃ (z=x v z=y))))
- (∀x) (B1x ⊃ (∃y) (T1y & H2xy & (∃z) (T1z & H2xz & y≠z & (∀x1) ((T1x1 & H2xx1) ⊃ (x1=y v x1=z)))))